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LatineSummary
This lesson opens the unit on derivatives. It presents the path through the unit, which defines the derivative as the limit of the difference quotient, writes it as a factorization with a continuous factor, Carathéodory’s, and obtains from it, without ever dividing by quantities that may vanish, the differentiation rules, the chain rule and the derivative of the inverse function, until it reaches higher-order derivatives, the differential and rates of change. It reconstructs the history of the derivative, from Fermat’s method for tangents to Cauchy’s definition, by way of Newton’s fluxions, Leibniz’s notation, Berkeley’s criticism and Lagrange’s program, which gave the derivative its name and its notation; it states the problem that gave rise to it, the tangent to an arbitrary curve and the velocity at an instant, and shows with an example how the factorization replaces division by a zero increment. It ends by pointing out where derivatives are used in the following unit, in other courses of the site and in science and technology.
Learning objectives
- Describe the path through the unit and explain what each lesson proves and which earlier results it relies on.
- Explain why the tangent to an arbitrary curve and instantaneous velocity require defining the derivative as a limit, and why Carathéodory’s factorization makes it possible to prove the rules without dividing by quantities that may vanish.
- Place in their historical context the stages of the notion of derivative, from Fermat’s method to Cauchy’s definition, and identify the problem or the criticism that drove each one.
- Point out where derivatives come into play in the following unit, in other courses and in scientific and technical applications.
Entry assessment
Before working through the unit, it is advisable to measure the starting point. This self-assessment is not graded: it serves to know where to pause. The first nine questions cover, in order, the nine content lessons of the unit, one each, and measure a skill of computation or of reasoning, not a statement to be remembered; the same skills are measured again at the end, in the synthesis lesson, with other data. The last five check what this unit takes as known from the preceding unit and from Algebra and Geometry I and II: a trigonometric limit, a standard limit of the exponential, little , the inverse function and the cosine of the arcsine.
Before starting the unit, measure your starting point. It is a short self-assessment: it is not graded and its result is stored only in your browser.
Question 1The function is defined at . Using the definition of the derivative, what is the tangent line to its graph at the point ?
This is the normal line, not the tangent: its slope, , makes it perpendicular to the tangent.The slope is correct, but the line does not pass through the point: the tangent is , and has been omitted.The sign has been lost in the subtraction: , which is negative for small . The function decreases near , and the slope is negative.For close to , the difference quotient is , which tends to ; therefore , and the tangent is the line through with that slope.Question 2Compute the derivative of at .
The numerator and the denominator have been differentiated separately, , as if the derivative of a quotient were the quotient of the derivatives. It is not: the quotient rule has two terms and the square of the denominator.The order of the subtraction in the numerator has been reversed. The rule is : first the derivative of the numerator times the denominator.The denominator has not been squared: at it equals , and its square is .By the quotient rule, .Review: Differentiation rules
Question 3Compute the derivative of at .
has been taken as the derivative of the cosine. The derivative of the cosine is , and the two terms cancel at .The derivatives of the two factors have been multiplied, . The derivative of a product is not the product of the derivatives: the term is missing.Only the exponential has been differentiated, as if the cosine were constant. That is the first term of the product rule; the second, , is missing.By the product rule and the derivatives of the exponential and of the cosine, , and at the sine and the cosine are equal. The tangent to the graph at that point is horizontal.Question 4Compute the derivative of at .
The derivative of the innermost function, , which contributes the factor , is missing: each function of the composition contributes its derivative, evaluated at the point that corresponds to it.has been taken as the derivative of the cosine: the minus sign of has been lost.Only the square has been differentiated, , as if the cosine were the variable. The derivatives of the cosine and of are missing.By the chain rule applied twice, , and at the argument is .Review: The chain rule for the derivative of a composition of functions
Question 5On which of these intervals does Bolzano’s theorem guarantee that the equation has a solution?
On that interval does not change sign: and . The theorem guarantees nothing, and in fact there is no solution there, because is increasing and equals at .and have the same sign, so the theorem does not apply; moreover, is at least on that interval.Every equation can be written in that form by moving everything to one side: is equivalent to .The function is a polynomial, hence continuous, with and ; by Bolzano’s theorem it vanishes at some point of , which is a solution of the equation.Review: The intermediate value theorem
Question 6The function is continuous and strictly increasing on . What is the value of ?
The derivative of has been evaluated at , which is a value of , not a point of its domain corresponding to it. The formula is , and .This is the derivative of at the point corresponding to ; that of the inverse is its reciprocal: when the graph is reflected, a slope becomes .has been computed, which is neither the derivative of the inverse nor evaluated at the correct point.Since , we have , and , which is not zero. By the theorem on the derivative of the inverse, .Review: Derivative of the inverse function and implicit differentiation
Question 7Compute the derivative of at .
The power rule has been used, , as if the exponent were constant. When both the base and the exponent vary, the power must be written with the natural exponential.The exponential rule has been used, , as if the base were constant. It is only one of the two terms.This is the derivative of , that is, the logarithmic derivative ; it still has to be multiplied by .For every , , and . At , and the parenthesis equals .Review: The variable in the exponent and logarithmic differentiation
Question 8Compute the derivative of order of at .
The binomial coefficient is missing: the Leibniz rule has the same coefficients as the binomial theorem.The derivatives of order of the two factors have been multiplied, and that of is zero. The derivative of order of a product is a sum over all the ways of distributing the five differentiations among the factors.In the term that differentiates the factor once, the exponential is differentiated four times, not five: its derivative of order is .By the Leibniz rule, since the derivatives of of order greater than are zero, only two terms remain: . At , the first vanishes and the second equals .Question 9A rectangle keeps its area of square centimeters while its base grows at a rate of centimeters per second. How fast is its height changing at the instant when the base measures centimeters?
has been substituted before differentiating. The height equals only at that instant; the relation to be differentiated is the one that holds at every instant.The sign has been lost: if the area is constant and the base grows, the height has to decrease.This is the derivative of the height with respect to the base, . The rate with respect to time is obtained by multiplying it by .If is the base and the height, at every instant; differentiating with respect to time, . When the base measures , the height measures , and gives .Question 10From earlier unitsCompute .
That is the limit of , not of this quotient: the cosine moves away from much more slowly than the sine moves away from .That is the limit of , with in the denominator. With , the quotient has one more factor , which tends to .only indicates that the quotient rule does not apply; the limit may exist, and here it does.For , the quotient is , the product of a function that tends to and another that tends to .Question 11From earlier unitsCompute .
The standard limit has been applied without changing the variable: the exponent is , not .The sign of the exponent has been lost: is less than for , so the quotient is negative.That the numerator tends to is not enough: the denominator also tends to .With the change of variable , which does not vanish if , the quotient is , and the standard limit gives .Review: Continuity of the elementary functions and standard limits
Question 12From earlier unitsWhich of these functions is as ?
It is equivalent to , not negligible: its quotient by tends to .It tends to , but more slowly than : its quotient by is not bounded.The summand is negligible compared with , but is not: the quotient by tends to , and the function is of the same order as .Its quotient by is , which tends to : it is negligible compared with .Review: Equivalent infinitesimals and infinitely large functions
Question 13From earlier unitsIf , what is the value of ?
has been isolated, but the cube root still has to be taken.has been read as the reciprocal : the exponent denotes here the inverse function, not the multiplicative inverse.This is , the image of ; the inverse travels the path in the opposite direction and looks for the preimage.is the number whose image is : gives , and .Review: Function operations, inverse functions and graph transformations
Question 14From earlier unitsWhat is the value of ?
The sign of the sine does not carry over to the cosine: the angle lies in , where the cosine is positive or zero.That is the square of the cosine, ; the square root still has to be taken.The cosine has been treated as if it undid the arcsine; it is the sine that undoes it: .The arcsine takes values in , where the cosine is not negative; by the Pythagorean identity, the cosine is .
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