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Topos Uranos
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Derivative of the inverse function and implicit differentiation

Summary

This lesson proves that the inverse of a function that is continuous and strictly monotonic on an interval is differentiable at every point at which the function has a nonzero derivative, and that its derivative is the reciprocal of the derivative of the function, evaluated at the corresponding point: when the graph is reflected about the line y=xy = x, a tangent of slope mm becomes one of slope 1m\frac{1}{m}. The proof uses the continuity of the inverse, from the lesson on the intermediate value theorem, and the Carathéodory characterization; where the derivative of the function vanishes, the inverse is not differentiable. With the theorem, the arcsine, the arccosine, the arctangent, the inverses of the hyperbolic functions, the roots and the powers with rational exponents are differentiated. The second part of the lesson differentiates functions defined by an equation, such as the branches of a circle or of the folium of Descartes, assuming that the function exists and is differentiable, and shows what happens when that assumption fails.

Learning objectives

  1. Prove the theorem on the derivative of the inverse function, and explain why the inverse is not differentiable where the derivative of the function vanishes.
  2. Differentiate the inverse trigonometric functions, the inverses of the hyperbolic functions, the roots and the powers with rational exponents, together with the set on which each derivative exists.
  3. Differentiate implicitly a function defined by an equation, assuming that it exists and is differentiable, and thus find the tangents to curves that are not graphs of functions.
  4. Recognize when implicit differentiation decides nothing: equations that define no differentiable function, and points at which the coefficient of the derivative vanishes.

The derivative of the inverse function

The graph of the inverse of an injective function is the reflection of the graph of the function about the line y=xy = x: the point (a,b)(a, b) of the former becomes the point (b,a)(b, a) of the latter. The reflection swaps the horizontal and the vertical increments, so a secant of slope mm becomes a secant of slope 1m\frac{1}{m}, and the same is to be expected of the tangents. The lesson on the derivatives of the elementary functions observed it in one case: the tangents to the graphs of exe^x and of ln⁡x\ln x at points symmetric about the line y=xy = x have reciprocal slopes, eae^a and 1ea\frac{1}{e^a}.

A quick computation suggests the general formula. If f−1f^{-1} were differentiable, differentiating the equality f(f−1(y))=yf\left(f^{-1}(y)\right) = y with the chain rule would give f′(f−1(y))(f−1)′(y)=1f'\left(f^{-1}(y)\right) \left(f^{-1}\right)'(y) = 1, and it would suffice to divide. But that computation assumes what has to be proved, that the inverse is differentiable, and says nothing about what happens if f′(f−1(y))=0f'\left(f^{-1}(y)\right) = 0. The case does occur: x3x^3 is differentiable at 00, with zero derivative, and its inverse, the cube root, is not differentiable at 00, by the proposition on continuity without a derivative, from the lesson on the derivative as a limit. The following theorem proves differentiability without assuming it, by means of the Carathéodory characterization, which makes it possible to isolate the increment of the inverse without dividing by any increment.

TheoremDerivative of the inverse function

Let ff be a continuous and strictly monotonic function on an interval II with more than one point, and let J=f(I)J = f(I). If ff is differentiable at a point aa of II and f′(a)≠0f'(a) \neq 0, then the inverse f−1 ⁣:J→If^{-1} \colon J \to I is differentiable at b=f(a)b = f(a), and

(f−1)′(b)=1f′(a)=1f′(f−1(b))\left(f^{-1}\right)'(b) = \frac{1}{f'(a)} = \frac{1}{f'\left(f^{-1}(b)\right)}

The Carathéodory factorization of ff at aa is read backwards: the function φ\varphi does not vanish, and its reciprocal, evaluated at f−1(y)f^{-1}(y), factors the increment of the inverse. The figure shows, with f(x)=x3f(x) = x^3 and a=0.8a = 0.8, a secant and the tangent of ff, and then their reflections, which are a secant and the tangent of the cube root.

Proof

  1. f(x)−f(a)=φ(x)(x−a),x∈If(x) - f(a) = \dato{fi}{\varphi(x)(x - a)}, \qquad x \in I

    Since II is an interval with more than one point, all its points are accumulation points of II. By the theorem on the characterizations of the derivative, from the lesson on the derivative as a limit, there exists φ ⁣:I→R\varphi \colon I \to \mathbb{R}, continuous at aa, with φ(a)=f′(a)\varphi(a) = f'(a), such that:

  2. x≠a→φ(x)=f(x)−f(a)x−a≠0,φ(a)=f′(a)≠0x \neq a \rightarrow \varphi(x) = \frac{f(x) - f(a)}{x - a} \neq 0, \qquad \resaltar{\varphi(a) = f'(a) \neq 0}

    The function φ\varphi does not vanish at any point of II. At aa its value is f′(a)f'(a), which is not zero; if x≠ax \neq a, strict monotonicity gives f(x)≠f(a)f(x) \neq f(a), and φ(x)\varphi(x) is the difference quotient, which is not zero either.

  3. f−1(y)−f−1(b)=1φ(f−1(y))(y−b),y∈Jf^{-1}(y) - f^{-1}(b) = \dato{psi}{\frac{1}{\varphi\left(f^{-1}(y)\right)}}(y - b), \qquad y \in J

    Let yy be in JJ, and let x=f−1(y)x = f^{-1}(y), so that f(x)=yf(x) = y and x−a=f−1(y)−f−1(b)x - a = f^{-1}(y) - f^{-1}(b). The factorization is divided by φ(x)\varphi(x), which is not zero: the increment of the inverse is factored, and there has been no division by any increment.

  4. ψ(b)=1φ(a)=1f′(a)\psi(b) = \frac{1}{\varphi(a)} = \resaltar{\frac{1}{f'(a)}}

    By the corollaries on the image of an interval and on the continuity of the inverse, from the lesson on the intermediate value theorem, JJ is an interval, with more than one point because ff is injective, and f−1f^{-1} is continuous at bb. Since φ\varphi is continuous at a=f−1(b)a = f^{-1}(b), by the theorems on the composition of continuous functions and on operations with continuous functions, from the lesson on limit and continuity, the factor ψ=1φ∘f−1\psi = \frac{1}{\varphi \circ f^{-1}} is continuous at bb. Since bb is an accumulation point of JJ, the Carathéodory characterization gives the derivative.

The continuity of ff on all of II is required so that the inverse is continuous at bb, which is what allows passing to the limit in the last move of the proof. If ff is differentiable on all of II and its derivative does not vanish, the inverse is differentiable on all of JJ, and the equality is written between functions: (f−1)′=1f′∘f−1\left(f^{-1}\right)' = \frac{1}{f' \circ f^{-1}}. The derivative of ff is evaluated at f−1(y)f^{-1}(y), not at yy. In Leibniz notation, if y=f(x)y = f(x) and x=f−1(y)x = f^{-1}(y), the theorem says that dxdy=1dydx\frac{\mathrm{d}x}{\mathrm{d}y} = \frac{1}{\frac{\mathrm{d}y}{\mathrm{d}x}}, with the derivative in the denominator evaluated at x=f−1(y)x = f^{-1}(y). As with the chain rule, the formula looks like a rule for fractions and is not one. Once differentiability has been proved, the quick computation at the beginning, which differentiates f(f−1(y))=yf\left(f^{-1}(y)\right) = y with the chain rule, is a legitimate way of remembering the formula.

PropositionThe inverse where the derivative vanishes

Let ff be a continuous and strictly monotonic function on an interval II with more than one point. If ff is differentiable at a point aa of II and f′(a)=0f'(a) = 0, then f−1f^{-1} is not differentiable at b=f(a)b = f(a). If it were, the chain rule, applied to the identity f−1(f(x))=xf^{-1}(f(x)) = x, would give 1=01 = 0.

Proof

  1. (f−1∘f)′(a)=(f−1)′(b) f′(a)=0\left(f^{-1} \circ f\right)'(a) = \left(f^{-1}\right)'(b) \, f'(a) = \resaltar{0}

    Suppose that f−1f^{-1} is differentiable at bb. The function ff is defined on II and is differentiable at aa, which is an accumulation point of II, and its values lie in J=f(I)J = f(I), where f−1f^{-1} is defined. By the chain rule, from the lesson on the chain rule:

  2. f−1(f(x))=x→(f−1∘f)′(a)=1≠0f^{-1}(f(x)) = x \rightarrow \left(f^{-1} \circ f\right)'(a) = \resaltar{1 \neq 0}

    But f−1∘ff^{-1} \circ f is the identity on II, whose derivative at every point is 11, as was seen in the lesson on the derivative as a limit. The contradiction shows that f−1f^{-1} is not differentiable at bb.

More can be said. If y≠by \neq b, the third move of the proof of the theorem only uses that φ\varphi does not vanish away from aa, and it holds even if f′(a)=0f'(a) = 0: the difference quotient of the inverse is 1φ(f−1(y))\frac{1}{\varphi\left(f^{-1}(y)\right)}, and the denominator tends to φ(a)=0\varphi(a) = 0, by the continuity of the composition. If ff is increasing, the denominator is positive, and the quotient tends to ∞\infty; if it is decreasing, it tends to −∞-\infty. The graph of the inverse has at (b,a)(b, a) a vertical tangent, the reflection of the horizontal tangent of ff at (a,b)(a, b), like the cube root at the origin.