{"id":35171,"date":"2024-12-01T13:00:07","date_gmt":"2024-12-01T13:00:07","guid":{"rendered":"https:\/\/toposuranos.com\/material\/?p=35171"},"modified":"2025-11-22T22:33:22","modified_gmt":"2025-11-22T22:33:22","slug":"chain-rule-for-the-derivative-of-the-composition-of-functions","status":"publish","type":"post","link":"https:\/\/toposuranos.com\/material\/en\/chain-rule-for-the-derivative-of-the-composition-of-functions\/","title":{"rendered":"Chain Rule for the Derivative of the Composition of Functions"},"content":{"rendered":"<style>\np, ul, ol{\ntext-align: justify;\n}\nh1{\ntext-align:center;\ntext-transform: uppercase;\n}\nh2{\ntext-align:center;\ntext-transform: uppercase;\nfont-size:24pt;\n}\nh3 { \n    text-align: center;\n    text-transform: uppercase;\n    font-size: 24px !important;\n}\n<\/style>\n<h1>Chain Rule for the Derivative of the Composition of Functions<\/h1>\n<p><em>With what we have seen so far, we already have all the basics to compute almost any derivative. However, we must distinguish between the possibility of computing a derivative and the effort we invest in carrying out such calculations, and this is where theorems such as the chain rule for the calculation of one variable come into play. The chain rule will allow us to quickly compute derivatives that would otherwise involve rather tedious and complicated work.<\/em><\/p>\n<p style=\"text-align:center;\" dir=\"ltr\">\n<b><u>TABLE OF CONTENTS<\/u><\/b><br \/>\n<b><a href=\"#1\">The Chain Rule Theorem in One Real Variable<\/a><\/b><br \/>\n<a href=\"#11\">Proof of the Chain Rule<\/a><br \/>\n<a href=\"#12\">Examples of the Use of the Chain Rule in Functions of One Variable<\/a><br \/>\n<a href=\"#13\">Precaution to Keep in Mind Regarding the Chain Rule<\/a><br \/>\n<b><a href=\"#2\">Useful Results Obtained from the Chain Rule<\/a><\/b><br \/>\n<a href=\"#21\">Inverse Function Theorem<\/a><br \/>\n<a href=\"#211\">Derivative of the Exponential Function<\/a><br \/>\n<a href=\"#212\">Derivative of the Inverse Trigonometric Functions<\/a><br \/>\n<a href=\"#22\">Implicit Differentiation<\/a><br \/>\n<a href=\"#221\">Derivatives of Rational Powers<\/a><br \/>\n<a href=\"#221\">Derivatives of Rational Powers<\/a><br \/>\n<b><a href=\"#3\">Exercise Guide<\/a><\/b>\n<\/p>\n<p><a name=\"1\"><\/a><br \/>\n<center><iframe class=\"lazyload\" width=\"560\" height=\"315\" data-src=\"https:\/\/www.youtube.com\/embed\/0y2SQpbRe3A\" title=\"YouTube video player\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center><\/p>\n<p><a name=\"1\"><\/a><\/p>\n<h2>The Chain Rule Theorem in One Real Variable<\/h2>\n<p><a href=\"https:\/\/www.youtube.com\/watch?v=0y2SQpbRe3A&amp;t=165s\" target=\"_blank\" rel=\"noopener\"><span style=\"color: #ff0000;\">Let <span class=\"katex-eq\" data-katex-display=\"false\">f<\/span> and <span class=\"katex-eq\" data-katex-display=\"false\">g<\/span> be two functions<\/span><\/a> that can be composed<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">f: A\\subseteq \\mathbb{R} \\longmapsto B\\subseteq \\mathbb{R}<\/span>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">g: B\\subseteq Dom(g) \\longmapsto D\\subseteq \\mathbb{R}<\/span>\n<p>If <span class=\"katex-eq\" data-katex-display=\"false\">f<\/span> is differentiable on <span class=\"katex-eq\" data-katex-display=\"false\">A<\/span> and <span class=\"katex-eq\" data-katex-display=\"false\">g<\/span> is differentiable on <span class=\"katex-eq\" data-katex-display=\"false\">B<\/span>, then the composite function <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">g\\circ f<\/span><\/span> is differentiable for all <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x\\in A<\/span><\/span> and the following formula holds<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\frac{d}{dx}(g\\circ f)(x) = \\frac{d}{dx} g(f(x)) = \\frac{dg(f(x))}{df(x)} \\frac{df(x)}{dx}<\/span>\n<p><a name=\"11\"><\/a><\/p>\n<h3>Proof of the Chain Rule<\/h3>\n<p><a href=\"https:\/\/www.youtube.com\/watch?v=0y2SQpbRe3A&amp;t=242s\" target=\"_blank\" rel=\"noopener\"><span style=\"color: #ff0000;\">Let us consider the functions<\/span><\/a> <span class=\"katex-eq\" data-katex-display=\"false\">f<\/span> and <span class=\"katex-eq\" data-katex-display=\"false\">g<\/span> as defined above. If we compute the derivative of the composition, then we have<\/p>\n<p style=\"text-align:center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\n\\begin{array}{rcl}\n\n\\dfrac{d}{dx} g(f(x))&amp; = &amp; \\displaystyle\\lim_{\\Delta x \\to 0} \\dfrac{g(f(x + \\Delta x)) - g(f(x))}{\\Delta x} \\\\ \\\\\n\n&amp;=&amp;\\displaystyle \\lim_{\\Delta x \\to 0} \\frac{g(f(x + \\Delta x)) - g(f(x))}{\\Delta x} \\cdot \\frac{f(x + \\Delta x) - f(x)}{f(x+\\Delta x) - f(x)} \\\\ \\\\\n\n&amp;=&amp; \\displaystyle \\lim_{\\Delta x \\to 0} \\frac{g(f(x + \\Delta x)) - g(f(x))}{f(x+\\Delta x) - f(x)} \\cdot \\frac{f(x + \\Delta x) - f(x)}{\\Delta x} \\\\ \\\\\n\n&amp;=&amp;\\displaystyle \\lim_{\\Delta x \\to 0} \\frac{g(f(x + \\Delta x)) - g(f(x))}{f(x+\\Delta x) - f(x)} \\cdot \\lim_{\\Delta x \\to 0} \\frac{f(x + \\Delta x) - f(x)}{\\Delta x}\\\\ \\\\\n\n&amp;=&amp; \\displaystyle \\lim_{f(x+\\Delta x) \\to f(x) } \\frac{g(f(x + \\Delta x)) - g(f(x))}{f(x+\\Delta x) - f(x)} \\cdot \\lim_{\\Delta x \\to 0} \\frac{f(x + \\Delta x) - f(x)}{\\Delta x}\\\\ \\\\\n\n&amp;=&amp; \\displaystyle \\frac{dg(f(x))}{df(x)} \\frac{df(x)}{dx}\n\n\\end{array}\n\n<\/span>\n<p>Which is what we wanted to prove.<\/p>\n<p><a name=\"12\"><\/a><\/p>\n<h3>Examples of the Use of the Chain Rule in Functions of One Variable<\/h3>\n<p><a href=\"https:\/\/www.youtube.com\/watch?v=0y2SQpbRe3A&amp;t=423s\" target=\"_blank\" rel=\"noopener\"><span style=\"color: #ff0000;\">Something that becomes clear, at least at first glance,<\/span><\/a> but is not so evident from an operational perspective, is the fact that the chain rule tells us that when we encounter a composition of functions, we can differentiate \u201cfrom the outside in.\u201d To explain this in a way that is easy to understand, examples are by far the quickest route.<\/p>\n<ol>\n<li>If we are asked to differentiate <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">f(x) = (2x^2+1)^{12}<\/span><\/span> we would first expand the powers and then apply the power rule to each part of the large polynomial we would have obtained as a result\u2014an unnecessarily exhausting task. With the chain rule, the computation of the derivative can be done in just a few lines:<br \/>\n<\/p>\n<p style=\"align:center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\frac{d}{dx} (2x^2+1)^{12} = 12(2x^2+1)^{11}(4x)= 48x(2x^2+1)^{11}<\/span>\n<\/li>\n<li>Try to compute the derivative of <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">g(x) = \\sin(\\cos(x))<\/span><\/span> using only basic differentiation techniques and face eternal suffering. Do it using the chain rule and the result will appear without tears and in just a few steps:<br \/>\n<\/p>\n<p style=\"align:center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\frac{d}{dx} \\sin(\\cos(x))= -\\cos(cos(x))\\sin(x) <\/span>\n<\/li>\n<li>You can also compute the derivative of functions that are compositions of many functions. If <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">f(x)=\\cos(\\cos(\\cos(x))),<\/span><\/span> the derivative <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">df\/dx<\/span><\/span> becomes:<br \/>\n<\/p>\n<p style=\"text-align:center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\n\\begin{array}{rcl}\n\n\\displaystyle \\frac{d}{dx} \\cos(\\cos(\\cos(x))) &amp;=&amp; -\\sin(\\cos(\\cos(x)))\\cdot(-\\sin(\\cos(x))\\cdot(-\\sin(x)) \\\\ \\\\\n\n&amp;=&amp; -\\sin(\\cos(\\cos(x)))\\cdot\\sin(\\cos(x))\\cdot\\sin(x)\n\n\\end{array}\n\n<\/span>\n<p>As you can see, applying the chain rule simply means differentiating in a chained manner from the outside in.<\/li>\n<\/ol>\n<p><a name=\"13\"><\/a><\/p>\n<h3>Precaution to Keep in Mind Regarding the Chain Rule<\/h3>\n<p><a href=\"https:\/\/www.youtube.com\/watch?v=0y2SQpbRe3A&amp;t=607s\" target=\"_blank\" rel=\"noopener\"><span style=\"color: #ff0000;\">In the literature, everyone shows the great benefits<\/span><\/a> of using the chain rule, but very few are emphatic about the precautions that must be taken before applying it. Despite the power of this theorem, you must always pay close attention to the domains and ranges of the functions before applying the chain rule. Before working, you must ensure that the domains and ranges of the functions are compatible for composition; otherwise, you run the risk of computing derivatives where they do not exist. If you differentiate, for example, a function of the form<\/p>\n<p style=\"text-align:center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">f(x)=\\ln(\\cos(x))<\/span>\n<p>if you blindly trust the chain rule, you will perform calculations such as the following:<\/p>\n<p style=\"text-align:center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\frac{d}{dx}\\ln(\\cos(x)) = -\\frac{1}{\\cos(x)}\\sin(x) = -\\tan(x)<\/span>\n<p>Clearly, the tangent function is well defined for the value <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x=2\\pi\/3<\/span><\/span>, because its value is <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\tan(2\\pi\/3) = -\\sqrt{3}<\/span><\/span>. However, the function <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">f(x)=\\ln(\\cos(x))<\/span><\/span> is not well defined there because <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">f(2\\pi\/3) = \\ln(\\cos(2\\pi\/3)) = \\ln(-1\/2),<\/span><\/span> and the logarithm of negative numbers does not exist! In cases like this, it is necessary to indicate, before applying the chain rule, that the values of <span class=\"katex-eq\" data-katex-display=\"false\">x<\/span> to be considered are such that they keep the cosine function positive (so that compatibility under composition is ensured), and only then will the chain rule hold.<\/p>\n<p><a name=\"2\"><\/a><\/p>\n<h2>Useful Results Obtained from the Chain Rule<\/h2>\n<p>The chain rule is not only useful for achieving derivative calculations that would otherwise be unbearable; it is also useful for further expanding differentiation techniques to many other functions. Below we will review these techniques, their results, and their proofs.<\/p>\n<p><a name=\"21\"><\/a><\/p>\n<p><center><iframe class=\"lazyload\" width=\"560\" height=\"315\" data-src=\"https:\/\/www.youtube.com\/embed\/5ddoUcIhgjU\" title=\"YouTube video player\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center><a name=\"21\"><\/a><\/p>\n<h3>Inverse Function Theorem<\/h3>\n<p><a href=\"https:\/\/www.youtube.com\/watch?v=5ddoUcIhgjU&amp;t=75s\" target=\"_blank\" rel=\"noopener\"><span style=\"color: #ff0000;\">Let <span class=\"katex-eq\" data-katex-display=\"false\">f<\/span> be a bijective function<\/span><\/a> and differentiable on some interval <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">I\\subseteq \\mathbb{R}<\/span><\/span>. Using the chain rule, it is possible to compute the derivative of the identity function <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">(f^{-1}\\circ f)(x) = f^{-1}(f(x)) = x.<\/span><\/span> The calculations yield the following result:<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">1 = \\displaystyle \\frac{d}{dx} x = \\frac{d}{dx} f^{-1}(f(x)) = \\frac{df^{-1}(f(x))}{df(x)}\\frac{df(x)}{dx}<\/span>\n<p>From this, one can solve for <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">df^{-1}(f(x))\/df(x)<\/span><\/span> and obtain:<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\color{blue}{\\frac{df^{-1}(f(x))}{df(x)}= \\frac{1}{\\frac{df(x)}{dx}}}<\/span>\n<p>This is what is known as the inverse function theorem for computing derivatives. In the literature, it is common to find this theorem written in the form<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\color{blue}{\\frac{dx}{dy}= \\frac{1}{\\frac{dy}{dx}}}<\/span>\n<p>Both ways of expressing the inverse function theorem are equivalent and follow from writing <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">y=f(x)<\/span><\/span> and <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x=f^{-1}(y).<\/span><\/span><\/p>\n<p>Up to this point, we have seen everything that can be said about what the inverse function theorem concerns; now we will see how we can use it to compute some derivatives that would otherwise be quite difficult.<\/p>\n<p><a name=\"211\"><\/a><\/p>\n<h4>Derivative of the Exponential Function<\/h4>\n<p><span style=\"color: #ff0000;\"><a href=\"https:\/\/www.youtube.com\/watch?v=5ddoUcIhgjU&amp;t=215s\" target=\"_blank\" style=\"color: #ff0000;\" rel=\"noopener\">When we studied basic differentiation techniques<\/a><\/span> we saw that<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\frac{d}{dx}\\ln(x) = \\frac{1}{x}<\/span>\n<p>With this result and the inverse function theorem, it is easy to prove that<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\frac{d}{dx}e^x = e^x<\/span>\n<p style=\"text-align: justify;color: #000080;\"><strong>PROOF:<\/strong><\/p>\n<p>It is clear that <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">y=\\ln(x)<\/span><\/span> is equivalent to saying that <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x=e^y.<\/span><\/span> Then, applying the inverse function theorem we have:<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\frac{d}{dy}e^y = \\frac{dx}{dy} = \\frac{1}{\\frac{dy}{dx}} = \\frac{1}{\\frac{d}{dx}\\ln(x)} = x = e^y<\/span>\n<p>That is:<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\frac{d}{dy}e^y = e^y<\/span>\n<p>If in this last expression we replace the \u00aby\u00bb with \u00abx\u00bb, we obtain what we wanted to prove:<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\frac{d}{dx}e^x = e^x.<\/span>\n<p><a name=\"212\"><\/a><\/p>\n<h4>Derivative of the Inverse Trigonometric Functions<\/h4>\n<p><a href=\"https:\/\/www.youtube.com\/watch?v=5ddoUcIhgjU\" target:=\"\" span=\"\" style=\"color: #ff0000;\" 0=\"\" a=\"\">The inverse function theorem<\/a> will also allow us to obtain the derivatives of all the inverse trigonometric functions. These are:<\/p>\n<p style=\"text-align:center;\" dir=\"ltr;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\begin{array}{ccccccc}\n\n\\dfrac{d}{dx}\\text{Arcsin}(x) &amp;=&amp; \\dfrac{1}{\\sqrt{1-x^2}} &amp;\\phantom{asd}&amp;\\dfrac{d}{dx}\\text{Arccos}(x) &amp;=&amp; \\dfrac{-1}{\\sqrt{1-x^2}} \\\\ \\\\\n\n\\dfrac{d}{dx}\\text{Arctan}(x) &amp;=&amp; \\dfrac{1}{1+x^2} &amp;\\phantom{asd}&amp;\\dfrac{d}{dx}\\text{Arccot}(x) &amp;=&amp; \\dfrac{-1}{1-x^2} \\\\ \\\\\n\n\\dfrac{d}{dx}\\text{Arcsec}(x) &amp;=&amp; \\dfrac{1}{x\\sqrt{x^2-1}} &amp;\\phantom{asd}&amp;\\dfrac{d}{dx}\\text{Arccsc}(x) &amp;=&amp; \\dfrac{-1}{x\\sqrt{x^2-1}}\n\n\\end{array}<\/span>\n<p style=\"text-align: justify; color: #000080;\"><strong>PROOF<\/strong><\/p>\n<h5>Arcsine<\/h5>\n<span class=\"collapseomatic \" id=\"id69e3f8fef32ae\"  tabindex=\"0\" title=\"Show Proof\"    >Show Proof<\/span><div id=\"target-id69e3f8fef32ae\" class=\"collapseomatic_content \">\n<p>The function <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\sin(x)<\/span><\/span> is bijective as long as we restrict its domain to a set of the form <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\left[\\frac{-\\pi}{2}+k\\pi , \\frac{\\pi}{2}+ k\\pi \\right],<\/span><\/span> with <span class=\"katex-eq\" data-katex-display=\"false\">k<\/span> any integer. Without loss of generality, it is possible to limit ourselves to the principal case, where <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">k=0<\/span><\/span>, so that the bijective sine function is of the form<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle\\sin : \\left[-\\frac{\\pi}{2}, \\frac{\\pi}{2}\\right] \\longrightarrow [-1,1]<\/span>\n<p>and under these conditions it holds that<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">y=\\sin(x) \\longleftrightarrow x=arcsin(y).<\/span>\n<p>If we apply the inverse function theorem, we have:<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\frac{d}{dy}arcsin(y) = \\frac{1}{\\frac{d}{dx}\\sin(x)} = \\frac{1}{\\cos(x)}<\/span>\n<p>Now, let us recall the trigonometric identity<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\sin^2(x) + \\cos^2(x) = 1<\/span>\n<p>from which it follows that, if <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x\\in [-\\pi\/2, \\pi\/2]<\/span><\/span>, then<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\"> \\cos(x) = \\sqrt{1 - \\sin^2(x)}<\/span>\n<p>Then, if we substitute this into the derivative of the arcsine function we obtain<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\frac{d}{dy}arcsin(y) = \\frac{1}{\\cos(x)} = \\frac{1}{ \\sqrt{1 - \\sin^2(x)}}<\/span>\n<p>And since <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">y=\\sin(x)<\/span><\/span><\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\frac{d}{dy}arcsin(y) = \\frac{1}{ \\sqrt{1 - y^2}}<\/span>\n<p>Finally, substituting \u201cy\u201d with \u201cx\u201d in this last expression, we arrive at what we wanted to prove:<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\color{blue}{\\frac{d}{dx}arcsin(x) = \\frac{1}{ \\sqrt{1 - x^2}}}<\/span>\n<\/div>\n<h5>Arccosine<\/h5>\n<span class=\"collapseomatic \" id=\"id69e3f8fef3487\"  tabindex=\"0\" title=\"Show Proof\"    >Show Proof<\/span><div id=\"target-id69e3f8fef3487\" class=\"collapseomatic_content \">\n<p>The function <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\cos(x)<\/span><\/span> is bijective as long as we restrict its domain to a set of the form <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\left[0+k\\pi , \\pi+ k\\pi \\right],<\/span><\/span> with <span class=\"katex-eq\" data-katex-display=\"false\">k<\/span> any integer. Without loss of generality, it is possible to limit ourselves to the principal case, where <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">k=0<\/span><\/span>, so that the bijective cosine function is of the form<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\cos : \\left[0, \\pi\\right] \\longrightarrow [-1,1]<\/span>\n<p>and under these conditions it holds that<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">y=\\cos(x) \\longleftrightarrow x=arccos(y).<\/span>\n<p>If we apply the inverse function theorem, we have:<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\frac{d}{dy}arccos(y) = \\frac{1}{\\frac{d}{dx}\\cos(x)} = \\frac{-1}{\\sin(x)}<\/span>\n<p>Now, let us recall the trigonometric identity<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\sin^2(x) + \\cos^2(x) = 1<\/span>\n<p>from which it follows that, if <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x\\in [0, \\pi]<\/span><\/span>, then<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\"> \\sin(x) = \\sqrt{1 - \\cos^2(x)}<\/span>\n<p>Then, if we substitute this into the derivative of the arccosine function, we obtain<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\frac{d}{dy}arccos(y) = \\frac{-1}{\\sin(x)} = \\frac{-1}{ \\sqrt{1 - \\cos^2(x)}}<\/span>\n<p>And since <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">y=\\cos(x)<\/span><\/span><\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\frac{d}{dy}arccos(y) = \\frac{-1}{ \\sqrt{1 - y^2}}<\/span>\n<p>Finally, substituting \u201cy\u201d with \u201cx\u201d in this last expression, we arrive at what we wanted to prove:<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\color{blue}{\\frac{d}{dx}arccos(x) = \\frac{-1}{ \\sqrt{1 - x^2}}}<\/span>\n<\/div>\n<h5>Arctangent<\/h5>\n<span class=\"collapseomatic \" id=\"id69e3f8fef360c\"  tabindex=\"0\" title=\"Show Proof\"    >Show Proof<\/span><div id=\"target-id69e3f8fef360c\" class=\"collapseomatic_content \">\n<p>The function <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\tan(x)<\/span><\/span> is bijective as long as we restrict its domain to a set of the form <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\left[-\\frac{\\pi}{2}+k\\pi , \\frac{\\pi}{2}+ k\\pi \\right],<\/span><\/span> with <span class=\"katex-eq\" data-katex-display=\"false\">k<\/span> any integer. Without loss of generality, it is possible to limit ourselves to the principal case, where <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">k=0,<\/span><\/span> so that the bijective tangent function is of the form<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\tan : \\left[-\\frac{\\pi}{2}, \\frac{\\pi}{2}\\right] \\longrightarrow \\mathbb{R}<\/span>\n<p>and under these conditions it holds that<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">y=\\tan(x) \\longleftrightarrow x=arctan(y).<\/span>\n<p>If we apply the inverse function theorem, we have:<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\frac{d}{dy}arctan(y) = \\frac{1}{\\frac{d}{dx}\\tan(x)} = \\frac{1}{\\sec^2(x)}<\/span>\n<p>Now, let us recall the trigonometric identity<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\sin^2(x) + \\cos^2(x) = 1<\/span>\n<p>from which it follows that<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\"> \\sec^2(x) =1+\\tan^2(x)<\/span>\n<p>Then, if we substitute this into the derivative of the arctangent function, we obtain<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\frac{d}{dy}arctan(y) = \\frac{1}{\\sec^2(x)} = \\frac{1}{ 1+\\tan^2(x)}<\/span>\n<p>And since <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">y=\\tan(x)<\/span><\/span><\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\frac{d}{dy}arctan(y) = \\frac{1}{1 + y^2}<\/span>\n<p>Finally, substituting \u201cy\u201d with \u201cx\u201d in this last expression, we arrive at what we wanted to prove:<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\color{blue}{\\frac{d}{dx}arctan(x) = \\frac{1}{1+ x^2}}<\/span>\n<\/div>\n<h5>Arccotangent<\/h5>\n<span class=\"collapseomatic \" id=\"id69e3f8fef3779\"  tabindex=\"0\" title=\"Show Proof\"    >Show Proof<\/span><div id=\"target-id69e3f8fef3779\" class=\"collapseomatic_content \">\n<p>The function <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">cot(x)<\/span><\/span> is bijective as long as we restrict its domain to a set of the form <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\left[0+k\\pi , \\pi+ k\\pi \\right],<\/span><\/span> with <span class=\"katex-eq\" data-katex-display=\"false\">k<\/span> any integer. Without loss of generality, it is possible to limit ourselves to the principal case, where <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">k=0<\/span><\/span>, so that the bijective cotangent function is of the form<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">ctg : \\left[0, \\pi\\right] \\longrightarrow \\mathbb{R}<\/span>\n<p>and under these conditions it holds that<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">y=ctg(x) \\longleftrightarrow x=arcctg(y).<\/span>\n<p>If we apply the inverse function theorem, we have:<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\frac{d}{dy}arcctg(y) = \\frac{1}{\\frac{d}{dx}ctg(x)} = \\frac{-1}{\\csc^2(x)}<\/span>\n<p>Now, let us recall the trigonometric identity<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\sin^2(x) + \\cos^2(x) = 1<\/span>\n<p>from which it follows that<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\"> \\csc^2(x) =1+ctg^2(x)<\/span>\n<p>Then, if we substitute this into the derivative of the arccotangent function, we obtain<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\frac{d}{dy}arcctg(y) = \\frac{-1}{\\csc^2(x)} = \\frac{-1}{ 1+ctg^2(x)}<\/span>\n<p>And since <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">y=ctg(x)<\/span><\/span><\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\frac{d}{dy}arcctg(y) = \\frac{-1}{1 + y^2}<\/span>\n<p>Finally, substituting \u201cy\u201d with \u201cx\u201d in this last expression, we arrive at what we wanted to prove:<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\color{blue}{\\frac{d}{dx}arcctg(x) = \\frac{-1}{1+ x^2}}<\/span>\n<\/div>\n<h5>Arcsecant<\/h5>\n<span class=\"collapseomatic \" id=\"id69e3f8fef38fd\"  tabindex=\"0\" title=\"Show Proof\"    >Show Proof<\/span><div id=\"target-id69e3f8fef38fd\" class=\"collapseomatic_content \">\n<p>The function <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\sec(x)<\/span><\/span> is bijective as long as we restrict its domain to a set of the form <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\left[0+k\\pi , \\pi+ k\\pi \\right]\\setminus\\left\\{\\frac{\\pi}{2} + k\\pi\\right\\},<\/span><\/span> with <span class=\"katex-eq\" data-katex-display=\"false\">k<\/span> any integer. Without loss of generality, we may limit ourselves to the principal case, where <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">k=0<\/span><\/span>, so that the bijective secant function is of the form<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\sec : \\left[0, \\pi\\right]\\setminus\\{\\pi\/2\\} \\longrightarrow \\mathbb{R}\\setminus]-1,1[<\/span>\n<p>and under these conditions it holds that<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">y=\\sec(x) \\longleftrightarrow x={arcsec}(y).<\/span>\n<p>If we apply the inverse function theorem, we have:<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\frac{d}{dy}{arcsec}(y) = \\frac{1}{\\frac{d}{dx}\\sec(x)} = \\frac{1}{\\sec(x)\\tan(x)}<\/span>\n<p>Now, let us recall the trigonometric identity<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\sin^2(x) + \\cos^2(x) = 1<\/span>\n<p>from which it follows that<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\"> \\tan^2(x) =\\sec^2(x)-1<\/span>\n<p>Then, if we substitute this into the derivative of the arcsecant function, we obtain<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\frac{d}{dy}{arcsec}(y) = \\frac{1}{\\sec(x)\\tan(x)} = \\frac{1}{sec(x)\\sqrt{\\sec^2(x)-1}}<\/span>\n<p>And since <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">y=\\sec(x)<\/span><\/span><\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\frac{d}{dy}{arcsec}(y) = \\frac{1}{y\\sqrt{y^2-1}}<\/span>\n<p>Finally, substituting \u201cy\u201d with \u201cx\u201d in this last expression, we arrive at what we wanted to prove:<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\color{blue}{\\frac{d}{dx}{arcsec}(x) = \\frac{1}{x\\sqrt{x^2-1}}}<\/span>\n<\/div>\n<h5>Arccosecant<\/h5>\n<span class=\"collapseomatic \" id=\"id69e3f8fef3a5e\"  tabindex=\"0\" title=\"Show Proof\"    >Show Proof<\/span><div id=\"target-id69e3f8fef3a5e\" class=\"collapseomatic_content \">\n<p>The function <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\csc(x)<\/span><\/span> is bijective as long as we restrict its domain to a set of the form <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\left[-\\frac{\\pi}{2}+k\\pi , \\frac{\\pi}{2} + k\\pi \\right]\\setminus\\left\\{0+k\\pi\\right\\}<\/span><\/span>, with <span class=\"katex-eq\" data-katex-display=\"false\">k<\/span> any integer. Without loss of generality, it is possible to limit ourselves to the principal case, where <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">k=0<\/span><\/span>, so that the bijective cosecant function is of the form<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\csc : \\left[-\\frac{\\pi}{2}, \\frac{\\pi}{2}\\right]\\setminus\\{0\\} \\longrightarrow \\mathbb{R}\\setminus]-1,1[<\/span>\n<p>and under these conditions it holds that<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">y=\\csc(x) \\longleftrightarrow x={arccsc}(y).<\/span>\n<p>If we apply the inverse function theorem, we have:<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\frac{d}{dy}{arccsc}(y) = \\frac{1}{\\frac{d}{dx}\\csc(x)} = \\frac{-1}{\\csc(x)ctg(x)}<\/span>\n<p>Now, let us recall the trigonometric identity<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\sin^2(x) + \\cos^2(x) = 1<\/span>\n<p>from which it follows that<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\"> ctg^2(x) =\\csc^2(x)-1<\/span>\n<p>Then, if we substitute this into the derivative of the arccosecant function, we obtain<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\frac{d}{dy}{arcsec}(y) = \\frac{-1}{\\csc(x)ctg(x)} = \\frac{-1}{csc(x)\\sqrt{\\csc^2(x)-1}}<\/span>\n<p>And since <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">y=\\csc(x)<\/span><\/span><\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\frac{d}{dy}{arccsc}(y) = \\frac{-1}{y\\sqrt{y^2-1}}<\/span>\n<p>Finally, substituting \u201cy\u201d with \u201cx\u201d in this last expression, we arrive at what we wanted to prove:<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\color{blue}{\\frac{d}{dx}{arccsc}(x) = \\frac{-1}{x\\sqrt{x^2-1}}}<\/span>\n<\/div>\n<p><a name=\"22\"><center><iframe class=\"lazyload\" width=\"560\" height=\"315\" data-src=\"https:\/\/www.youtube.com\/embed\/hOAydWcd6zw\" title=\"YouTube video player\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture\" allowfullscreen><\/iframe><\/center><\/a><\/p>\n<h3>Implicit Differentiation<\/h3>\n<p>All the derivatives we have computed up to now have been carried out on functions that were defined explicitly: <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">y=f(x)<\/span><\/span>. However, there are situations in which, based on the relationship between variables, it is either not easy to obtain the explicit expression of the function or such a task is simply not possible. For these types of cases, the technique of implicit differentiation is useful, and its foundations lie, once again, in the chain rule.<\/p>\n<p>To understand this technique, examples are more valuable than proofs, so let us consider the relationship between the variables <span class=\"katex-eq\" data-katex-display=\"false\">x<\/span> and <span class=\"katex-eq\" data-katex-display=\"false\">y<\/span> given by the equation<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x^3 +y^3- 9xy=0<\/span>\n<p>If we graph this relationship, we will realize that it is not the graph of any function. It is the graph of a curve called the \u201cDescartes\u2019 Folium.\u201d<\/p>\n<p><center><img decoding=\"async\" src=\"data:image\/gif;base64,R0lGODlhAQABAIAAAAAAAP\/\/\/yH5BAEAAAAALAAAAAABAAEAAAIBRAA7\" data-src=\"https:\/\/1.bp.blogspot.com\/-l30tAMcTkk0\/YLCIuWcDueI\/AAAAAAAAFIY\/K7uSR44DepgIjBlSVV7mCQO-Z0iy_RnRQCLcBGAsYHQ\/s0\/hojaDeDescartes.PNG\" alt=\"hoja de descartes\" class=\"alignnone size-full lazyload\" width=\"690\" height=\"515\" \/><noscript><img decoding=\"async\" src=\"https:\/\/1.bp.blogspot.com\/-l30tAMcTkk0\/YLCIuWcDueI\/AAAAAAAAFIY\/K7uSR44DepgIjBlSVV7mCQO-Z0iy_RnRQCLcBGAsYHQ\/s0\/hojaDeDescartes.PNG\" alt=\"hoja de descartes\" class=\"alignnone size-full lazyload\" width=\"690\" height=\"515\" \/><\/noscript><\/center><\/p>\n<p>Now, if we wanted to compute, for example, the derivative of <span class=\"katex-eq\" data-katex-display=\"false\">y<\/span> with respect to <span class=\"katex-eq\" data-katex-display=\"false\">x<\/span>, then we would face serious difficulties in finding an explicit expression <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">f(x)<\/span><\/span> that satisfies the equation <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">y=f(x)<\/span><\/span> in order to differentiate afterwards. What we do, however, is skip that step and implicitly assume that <span class=\"katex-eq\" data-katex-display=\"false\">y<\/span> is a function of <span class=\"katex-eq\" data-katex-display=\"false\">x<\/span>, that is: <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">y=y(x)<\/span><\/span>. Doing so transforms the Descartes\u2019 Folium relation into:<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x^3 +y^3(x)- 9xy(x)=0<\/span>\n<p>And we can consequently differentiate everything using the chain rule. If we do so, we arrive at the following result:<\/p>\n<p style=\"text-align:center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\n\\begin{array}{rcl}\n\n\\displaystyle 3x^{2} + 3\\,y(x)^{2}\\,\\frac{dy}{dx} - \\left(9\\,y(x) + 9x\\,\\frac{dy}{dx}\\right) &amp;=&amp; 0 \\\\ \\\\\n\n\\displaystyle 3x^{2} + 3\\,y(x)^{2}\\,\\frac{dy}{dx} - 9\\,y(x) - 9x\\,\\frac{dy}{dx} &amp;=&amp; 0 \\\\ \\\\\n\n\\displaystyle \\frac{dy}{dx}\\,\\big(3\\,y(x)^{2} - 9x\\big) &amp;=&amp; 9\\,y(x) - 3x^{2} \\\\ \\\\\n\n\\displaystyle \\frac{dy}{dx} &amp;=&amp; \\dfrac{9\\,y(x) - 3x^{2}}{3\\,y(x)^{2} - 9x} \\\\ \\\\\n\n\\displaystyle \\color{blue}{\\frac{dy}{dx}} &amp;\\color{blue}{=}&amp; \\color{blue}{\\dfrac{3\\,y(x) - x^{2}}{y(x)^{2} - 3x}}\n\n\\end{array}\n\n<\/span>\n<p>From this we can compute, if we know a point on the curve, the slope of the tangent line that passes through that point. For example, from the graph we can guess that the point <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">(2,4)<\/span><\/span> lies on the curve; and in fact, this is confirmed because <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">2^3 + 4^3 - 9\\cdot 2\\cdot 4 = 8+64 - 72 = 0. <\/span><\/span> Knowing this, we can quickly say that the slope of the tangent line that passes through that point will be:<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\color{blue}{\\left.\\frac{dy}{dx}\\right|_{(2,4)}= \\frac{3\\cdot 4 - 2^2}{4^2 - 3\\cdot 2}= \\frac{8}{10}= \\frac{4}{5}}<\/span>\n<p><a name=\"221\"><\/a><\/p>\n<h4>Derivatives of Rational Powers<\/h4>\n<p>By differentiating implicitly, it is possible to extend the reach of one of the basic differentiation techniques. This is the derivative of functions of the type <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">f(x)=x^n<\/span><\/span>, with <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n\\in\\mathbb{Z}<\/span><\/span>. We can now move from considering integers to rational numbers and show without difficulty that<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\frac{d}{dx}x^{p\/q}= \\frac{p}{q}x^{(p\/q) -1}<\/span>\n<p>where <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">p,q\\in\\mathbb{Z}<\/span><\/span> and <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">q\\neq 0<\/span><\/span>.<\/p>\n<p>To prove this, we say: let <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">y=x^{p\/q}<\/span><\/span> and apply the natural logarithm to obtain:<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\ln(y) = \\displaystyle \\frac{p}{q}\\ln(x)<\/span>\n<p>Now, differentiating this expression implicitly yields:<\/p>\n<p style=\"text-align:center;\" dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\n\\displaystyle \\frac{1}{y}\\frac{dy}{dx} = \\frac{p}{q}\\frac{1}{x}<\/span>\n<span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\color{blue}{\\frac{dy}{dx} = \\frac{p}{q}\\frac{1}{x}y(x)= \\frac{p}{q}\\frac{1}{x}x^{p\/q} = \\frac{p}{q}x^{(p\/q) - 1}}\n\n<\/span>\n<p><a name=\"3\"><center><iframe class=\"lazyload\" width=\"560\" height=\"315\" data-src=\"https:\/\/www.youtube.com\/embed\/KwJ5Bb5Ch_o\" title=\"YouTube video player\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture\" allowfullscreen><\/iframe><\/center><\/a><\/p>\n<h2>Exercise Guide:<\/h2>\n<h4>Chain Rule in One Variable<\/h4>\n<ol>\n<li>Compute the derivatives of the following group of functions:<br \/>\n<table>\n<tbody>\n<tr>\n<td width=\"20px\">a.<\/td>\n<td><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">f(x)=(x^2-3)^{12}<\/span><\/span><\/td>\n<td width=\"20px\">b.<\/td>\n<td><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">f(x)=\\displaystyle \\left(\\frac{4x^3 - x\\cos(2x) - 1}{\\sin(2x) + 2} \\right)^5<\/span><\/span><\/td>\n<\/tr>\n<tr>\n<td width=\"20px\">c.<\/td>\n<td><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">f(x)=\\cos(1-x^2)<\/span><\/span><\/td>\n<td width=\"20px\">d.<\/td>\n<td><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">f(x)=\\tan(x\\cos(3-x^2))<\/span><\/span><\/td>\n<\/tr>\n<tr>\n<td width=\"20px\">e.<\/td>\n<td><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">f(x)=\\displaystyle \\frac{1}{(\\sec(2x)-1)^{3\/2}}<\/span><\/span><\/td>\n<td width=\"20px\">f.<\/td>\n<td><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">f(x)=\\displaystyle \\frac{\\tan(2x)}{1-\\cot(2x)}<\/span><\/span><\/td>\n<\/tr>\n<tr>\n<td width=\"20px\">g.<\/td>\n<td><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">f(x)=\\displaystyle \\ln\\left(\\frac{\\tan(x)}{x^2+1}\\right)<\/span><\/span><\/td>\n<td width=\"20px\">h.<\/td>\n<td><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">f(x)=3^{\\csc(4x)}<\/span><\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/li>\n<li>Compute the derivative of the following group of functions:<br \/>\n<table>\n<tbody>\n<tr>\n<td width=\"20px\">a.<\/td>\n<td><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">f(x)=\\displaystyle \\frac{1}{\\sqrt{x}arctan\\left(x^3\\right)}<\/span><\/span><\/td>\n<td width=\"20px\">b.<\/td>\n<td><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">f(x)=\\displaystyle \\frac{{arcsec}(x^2-x+2)}{\\sqrt{x^2+1}}<\/span><\/span><\/td>\n<\/tr>\n<tr>\n<td width=\"20px\">c.<\/td>\n<td><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">f(x)=x^x<\/span><\/span><\/td>\n<td width=\"20px\">d.<\/td>\n<td><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">f(x)={arccsc}\\left(x^{\\ln(x)}\\right)<\/span><\/span><\/td>\n<\/tr>\n<tr>\n<td width=\"20px\">e.<\/td>\n<td><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">f(x)=\\ln\\left(arctan(e^x)\\right)<\/span><\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/li>\n<\/ol>\n","protected":false},"excerpt":{"rendered":"<p>Chain Rule for the Derivative of the Composition of Functions With what we have seen so far, we already have all the basics to compute almost any derivative. However, we must distinguish between the possibility of computing a derivative and the effort we invest in carrying out such calculations, and this is where theorems such [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":35164,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"iawp_total_views":2,"footnotes":""},"categories":[854,567],"tags":[],"class_list":["post-35171","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-differential-calculus","category-mathematics"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v26.7 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>Chain Rule for the Derivative of the Composition of Functions - toposuranos.com\/material<\/title>\n<meta name=\"description\" content=\"Chain rule in calculus: simple explanation, proofs, examples, and exercises. 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