Derivatives of Inverse Functions and Logarithms

What is the derivative of an inverse function?

If f has an interval as domain and f' exists and is never zero there, then f^-1 is differentiable at every point of its domain and \left(f^-1\right)'(b)=(1)/(f'\left(f^-1(b)\right)). In words: the slope of f^-1 at x=b is the reciprocal of the slope of f at x=f^-1(b). The inverse of f undoes it, so its graph is the mirror image of the graph of f in the line y=x, and a mirror turns a slope m into \frac1m.

A line first. The function f(x)=\frac12x+1 has inverse f^-1(x)=2x-2. Their derivatives are \frac12 and 2, reciprocals of each other. Reflecting any line that is neither horizontal nor vertical in y=x inverts its slope.

Other functions: compare at corresponding points. If the slope of y=f(x) at the point (a,b) is f'(a), with b=f(a), then the slope of y=f^-1(x) at the mirror point (b,a) is (1)/(f'(a)). The derivative of f is read at a and the derivative of f^-1 at b=f(a). If f has a horizontal tangent at (a,b), its inverse has a vertical tangent at (b,a) and no derivative there, which is why f'\ne0 is part of the condition.

Why the formula holds. Start from f\left(f^-1(x)\right)=x and differentiate both sides, using the chain rule on the left: f'\left(f^-1(x)\right)·\left(f^-1\right)'(x)=1. Solve for \left(f^-1\right)'(x) and the rule appears. The proof assumes that f^-1 is differentiable; that it is, when f'\ne0, is the first half of the theorem, which this article takes on trust.

Example 1. The function f(x)=x^2 for x>0 has the inverse f^-1(x)=√ x. The rule says \left(f^-1\right)'(x)=(1)/(f'\left(√ x\right))=(1)/(2√ x), the known derivative of the square root. At a=2 and b=f(2)=4: f'(2)=4 and \left(f^-1\right)'(4)=\frac14.

Example 2. Let f(x)=x^3-2 for x>0. Find \left(f^-1\right)'(6) without a formula for f^-1. First find a with f(a)=6: a^3-2=6, so a=2. Then f'(2)=3·2^2=12, and \left(f^-1\right)'(6)=(1)/(f'(2))=(1)/(12). Here it is the equation f(a)=6 that does the work of the missing formula.

Choose a function below and slide a: the mirror-image graph and the two tangent lines are drawn, and the slope of the inverse is checked against a numerical inverse.

What is the derivative of ln x?

The derivative of the natural logarithm is (d)/(dx)\ln x=\frac1x for x>0, and with the chain rule, (d)/(dx)\ln u=1/u·(du)/(dx) for a positive differentiable u(x). The function \ln x is the inverse of e^x, whose derivative is e^x itself. By the rule for inverses, with f(x)=e^x and f^-1(x)=\ln x: \left(f^-1\right)'(x)=(1)/(f'\left(\ln x\right))=(1)/(e^\ln x)=\frac1x.

A second derivation, by implicit differentiation. Let y=\ln x, so that e^y=x. Differentiate both sides with respect to x, using the chain rule on e^y: e^y (dy)/(dx)=1, (dy)/(dx)=(1)/(e^y)=\frac1x. This is the implicit differentiation of an equation that cannot be solved for y without the very function being differentiated.

Example 3. Four derivatives with \ln.

• (d)/(dx)\ln2x=(1)/(2x)·2=\frac1x. The constant disappears: \ln2x=\ln2+\ln x.
• (d)/(dx)\ln\left(x^2+3\right)=(1)/(x^2+3)·2x=(2x)/(x^2+3).
• (d)/(dx)\ln^3x=3\ln^2x·\frac1x=(3\ln^2x)/(x), since the outside is a cube and the inside is \ln x.
• (d)/(dx)\ln|x|=(1)/(|x|)·(x)/(|x|)=\frac1x for x\ne0, by (d)/(dx)|x|=(x)/(|x|). So \frac1x is the derivative of \ln x on x>0 and of \ln(-x) on x<0.

Logarithms of products and powers first. The laws of logarithms make many derivatives shorter. For instance \ln√(x^2+1)=\frac12\ln\left(x^2+1\right) has the derivative \frac12·(2x)/(x^2+1)=(x)/(x^2+1), with no square-root derivative at all.

Example 4. A line of slope m through the origin is tangent to the graph of y=\ln x. What is m? Let the point of tangency be (a,\ln a). The slope there is \frac1a, and a line through the origin and (a,\ln a) has slope (\ln a)/(a). Setting them equal, \frac1a=(\ln a)/(a), \ln a=1, a=e, m=\frac1e. The tangent line is y=x/e, and it touches the graph at (e,1).

The curve \ln x climbs ever more slowly, and the slope \frac1x falls from large values near 0 to small ones.

What are the derivatives of a^x and log base a, and where does e come from?

For a>0, (d)/(dx)a^x=a^x\ln a, and for a>0 with a\ne1, (d)/(dx)\log_ax=(1)/(x\ln a). Both follow from the natural logarithm. By definition a^x=e^x\ln a, so by the chain rule the derivative of the exponential e^u with u=x\ln a is e^x\ln a·\ln a=a^x\ln a. With a function u(x) in the exponent, (d)/(dx)a^u=a^u\ln a·(du)/(dx). For a logarithm, the change-of-base formula \log_ax=(\ln x)/(\ln a) shows a constant (1)/(\ln a) in front of \ln x, so (d)/(dx)\log_au=(1)/(u\ln a)·(du)/(dx).

Example 5. Derivatives of general exponentials.

• (d)/(dx)3^x=3^x\ln3.
• (d)/(dx)3^-x=3^-x\ln3·(-1)=-3^-x\ln3.
• (d)/(dx)3^\sin x=3^\sin x\ln3·\cos x.
• (d)/(dx)\sin\left(3^x\right)=\cos\left(3^x\right)·3^x\ln3, with two layers, the sine and the exponential.

Example 6. Derivatives of logarithms to other bases: (d)/(dx)\log_10x=(1)/(x\ln10), and (d)/(dx)\log_2\left(x^2+1\right)=(2x)/(\left(x^2+1\right)\ln2).

Why e is the preferred base. If a=e then \ln a=1 and the factor disappears: (d)/(dx)e^x=e^x, and (d)/(dx)\log_ex=\frac1x. For any other base a constant \ln a follows every derivative. The slope of a^x where its graph crosses the y-axis is a^0\ln a=\ln a, and by the definition of the derivative it is the limit \lim_h→0(a^h-1)/(h)=\ln a. For a=e that limit is 1: e is the base for which the curve a^x leaves the y-axis with slope exactly 1.

The number e as a limit. The derivative of \ln at 1 is \frac11=1, and by the definition of the derivative it is \lim_h→0(\ln(1+h)-\ln1)/(h)=\lim_h→0(\ln(1+h))/(h). So (\ln(1+h))/(h)→1. Since (\ln(1+h))/(h)=\ln\left((1+h)^1/h\right) and e^x is continuous, \lim_h→0(1+h)^1/h=e^1=e. This is the limit behind compound interest. With h=\frac1n it reads \left(1+\frac1n\right)^n→ e: money that earns interest 12 times a year grows by a factor \left(1+(1)/(12)\right)^12≈2.613 over a year at a nominal rate of 100\%, 365 times a year by 2.7146, and continuously by e≈2.71828.

An honest remark. The derivations above assume that the limits (a^h-1)/(h) exist. They do, and a later chapter of the book defines \ln and e^x in a way that proves it; here they are taken as known.

Shrink h below and watch the three limits settle.

What is logarithmic differentiation, and when does it help?

Logarithmic differentiation takes the natural logarithm of both sides of y=f(x), simplifies with the laws of logarithms, differentiates implicitly, and multiplies by y. It is quickest for products, quotients and powers of many factors, and it is the only method for a function with a variable exponent.

1. Take \ln of both sides and simplify: \ln(ab)=\ln a+\ln b, \ln\frac ab=\ln a-\ln b, \ln a^n=n\ln a.
2. Differentiate both sides with respect to x, which gives (y')/(y) on the left (see implicit differentiation).
3. Solve for y', and substitute y.

Example 7. Find y' for y=(x^2(x+1)^3)/(x-2), with x>2. Take logarithms: \ln y=2\ln x+3\ln(x+1)-\ln(x-2). Differentiate both sides: (y')/(y)=\frac2x+(3)/(x+1)-(1)/(x-2). Multiply by y: y'=(x^2(x+1)^3)/(x-2)\left(\frac2x+(3)/(x+1)-(1)/(x-2)\right). At x=3: y=(9·64)/(1)=576 and the bracket is \frac23+\frac34-1=(5)/(12), so y'(3)=576·(5)/(12)=240. The product and quotient rules would reach the same number, with a longer road.

Example 8: a variable exponent. For y=x^x with x>0 neither the power rule (the exponent is not constant) nor the rule for a^x (the base is not constant) applies. Take logarithms: \ln y=x\ln x. Differentiate with the product rule: (y')/(y)=\ln x+x·\frac1x=\ln x+1 y'=x^x(\ln x+1).

Example 9. For y=x^\sin x: \ln y=\sin x\ln x and (y')/(y)=\cos x\ln x+(\sin x)/(x), so y'=x^\sin x\left(\cos x\ln x+(\sin x)/(x)\right).

Negative values. The logarithm needs positive numbers. If a factor of y is negative, or y itself is, write \ln|y| instead: (d)/(dx)\ln|y|=(y')/(y) as well, so the same steps work.

Type a function below: the tool applies the rules, shows logarithmic differentiation for a product, a quotient or a power, and checks the result with a difference quotient.

Why does the power rule hold for irrational exponents?

For x>0 and any real number n, (d)/(dx)x^n=nx^n-1, because x^n is defined as e^n\ln x. Until now powers were defined by repeated multiplication and roots, which cannot raise a number to the power π or √2. The exponential and the logarithm can: for x>0 and every real n, x^n=e^n\ln x, so \ln x^n=n\ln x for all real n, the law of logarithms that was only proved for rational exponents.

Proof. Differentiate e^n\ln x with the chain rule: the outside is the exponential, which stays e^n\ln x=x^n, and the inside is n\ln x, with derivative \frac nx: (d)/(dx)x^n=x^n·\frac nx=nx^n-1.

Examples. (d)/(dx)x^π=π x^π-1 and (d)/(dx)x^√2=√2 x^√2-1.

Three different cases. The look of an expression with an exponent tells which rule to reach for.

The first two are single rules; the third is the case that needs the logarithm. A common error is to treat x^x with the power rule, as x· x^x-1, or with the exponential rule, as x^x\ln x: each gives only one of the two terms of x^x\ln x+x^x.

How do you differentiate logarithms and inverses in code?

SymPy differentiates sp.log and general powers directly: sp.diff(sp.log(x), x) is 1/x and `sp.diff(xx, x) is x**x*(log(x) + 1).** In SymPy log is the natural logarithm, and sp.log(x, 10) is the base-10 one, log(x)/log(10). Declaring x` positive lets it simplify logarithms of powers. The second program checks the rule for inverses without a formula for f^-1: it finds f^-1(6) for f(x)=x^3-2 by bisection, then compares (1)/(f'(a)) with a difference quotient of the numerical inverse. Both give (1)/(12)≈0.083333.

import sympy as sp

x = sp.symbols('x', positive=True)
for f in (sp.log(x), sp.log(2 * x), sp.log(x**2 + 3), sp.log(x, 10), 3**x, 3 ** sp.sin(x), x**x, x ** sp.sin(x), x**sp.pi):
    print(f, '->', sp.simplify(sp.diff(f, x)))
# log(x) -> 1/x
# log(2*x) -> 1/x
# log(x**2 + 3) -> 2*x/(x**2 + 3)
# log(x)/log(10) -> 1/(x*log(10))
# 3**x -> 3**x*log(3)
# 3**sin(x) -> 3**sin(x)*log(3)*cos(x)
# x**x -> x**x*(log(x) + 1)
# x**sin(x) -> x**(sin(x) - 1)*(x*log(x)*cos(x) + sin(x))
# x**pi -> pi*x**(-1 + pi)

y = x**2 * (x + 1) ** 3 / (x - 2)                    # Example 7
print(sp.diff(y, x).subs(x, 3), y.subs(x, 3), sp.diff(sp.log(y), x).subs(x, 3))   # 240 576 5/12

import math


def inverse(f, b, lo, hi):
    """f is increasing on [lo, hi]: find f^-1(b) by bisection."""
    for _ in range(200):
        mid = (lo + hi) / 2
        if f(mid) < b:
            lo = mid
        else:
            hi = mid
    return (lo + hi) / 2


f = lambda t: t**3 - 2
df = lambda t: 3 * t**2
h = 1e-6
a = inverse(f, 6, 0, 3)
print(round(a, 6), round(1 / df(a), 6))                                          # 2.0 0.083333
print(round((inverse(f, 6 + h, 0, 3) - inverse(f, 6 - h, 0, 3)) / (2 * h), 6))   # 0.083333

for h in (0.1, 0.01, 0.001, 0.0001, 0.000001):
    print(h, round((1 + h) ** (1 / h), 6), round(math.log(1 + h) / h, 6), round((2**h - 1) / h, 6))
# 0.1      2.593742  0.953102  0.717735
# 0.01     2.704814  0.995033  0.695555
# 0.001    2.716924  0.9995    0.693387
# 0.0001   2.718146  0.99995   0.693171
# 1e-06    2.71828   0.999999  0.693147
print(round(math.e, 6), round(math.log(2), 6))                                   # 2.718282 0.693147

#include <iostream>
#include <cstdio>
#include <cmath>
using namespace std;

double f(double t) { return t * t * t - 2; }

// f is increasing on [lo, hi]: find f^-1(b) by bisection
double inverse(double b, double lo, double hi) {
  for (int i = 0; i < 200; i++) {
    double mid = (lo + hi) / 2;
    if (f(mid) < b) lo = mid; else hi = mid;
  }
  return (lo + hi) / 2;
}

int main() {
  double h = 1e-6, a = inverse(6, 0, 3);
  printf("%.6f %.6f\n", a, 1 / (3 * a * a));                      // f^-1(6) = 2 and 1 / f'(2)
  printf("%.6f\n", (inverse(6 + h, 0, 3) - inverse(6 - h, 0, 3)) / (2 * h));   // the slope of the inverse

  double hs[5] = {0.1, 0.01, 0.001, 0.0001, 0.000001};
  for (int i = 0; i < 5; i++) {
    double x = hs[i];
    printf("%.6f  %.6f  %.6f  %.6f\n", x, pow(1 + x, 1 / x), log(1 + x) / x, (pow(2, x) - 1) / x);
  }
  printf("%.6f %.6f\n", exp(1.0), log(2.0));                      // e and ln 2
  return 0;
}

/* output:
2.000000 0.083333
0.083333
0.100000  2.593742  0.953102  0.717735
0.010000  2.704814  0.995033  0.695555
0.001000  2.716924  0.999500  0.693387
0.000100  2.718146  0.999950  0.693171
0.000001  2.718280  0.999999  0.693147
2.718282 0.693147
*/

Who invented logarithms, and when did e and the derivative of ln appear?

Logarithms came first as a way to multiply large numbers by adding, and the calculus of \ln was found when the area under the hyperbola was recognized as a logarithm.

• Napier and Briggs. The Scottish mathematician John Napier published his table of logarithms in 1614, to turn the long multiplications of astronomy into additions. Henry Briggs, working with Napier, made tables of base-10 logarithms, published from 1617.
• The hyperbola. In the 1640s Gregory of Saint-Vincent and his pupil de Sarasa showed that the area under y=\frac1x grows by equal steps when x grows by equal factors: the area is a logarithm. That is \ln x=\int_1^x(dt)/(t), and it is why its derivative is \frac1x.
• Mercator. Nicolaus Mercator named the natural logarithm and published its series \ln(1+x)=x-(x^2)/(2)+(x^3)/(3)-… in 1668.
• The number e. In 1683 Jacob Bernoulli studied the limit of \left(1+\frac1n\right)^n in a problem about compound interest. In the 1730s Leonhard Euler gave the number the letter e and developed the exponential and the logarithm as inverses of each other.

What mistakes do people make with inverses and logarithms?

The most common mistakes with derivatives of inverses and logarithms are the nine below, each with the correct statement.

Practice: test your understanding

These questions mix everything above. A wrong answer costs nothing here: read the hint and try again.

Summary: inverses and logarithms at a glance

• Inverses: \left(f^-1\right)'(b)=(1)/(f'\left(f^-1(b)\right)) when f'\ne0; slopes at mirror points are reciprocals.
• Natural logarithm: (d)/(dx)\ln x=\frac1x, (d)/(dx)\ln u=(u')/(u), (d)/(dx)\ln|x|=\frac1x.
• Other bases: (d)/(dx)a^u=a^u\ln a· u' and (d)/(dx)\log_au=(u')/(u\ln a); \ln e=1 makes e the natural base.
• The number e: \lim(1+h)^1/h=e and \lim(a^h-1)/(h)=\ln a.
• Logarithmic differentiation: \ln y, split, differentiate, multiply by y; (d)/(dx)x^x=x^x(\ln x+1).
• Power rule: (d)/(dx)x^n=nx^n-1 for every real n because x^n=e^n\ln x; x^π, π^x and x^x are three cases.
• Code: sp.diff(sp.log(x), x); check an inverse with bisection and a difference quotient.

Glossary: key terms

Inverse function

A function that undoes another, so that applying one after the other returns the starting number; its graph is the mirror image of the first in the line y equals x.

Natural logarithm

The logarithm to the base e, written ln x, the inverse of the exponential function e to the x, whose derivative is 1 over x.

Logarithmic differentiation

Finding a derivative by taking the logarithm of both sides, simplifying with the laws of logarithms, differentiating implicitly, and multiplying by the original function.

Euler’s number e

The number about 2.71828 that is the limit of one plus h to the power one over h as h tends to zero, and the base for which the slope of the exponential at the y-axis is exactly 1.

Change-of-base formula

The identity that log base a of x equals ln x divided by ln a, which reduces every logarithm to the natural one.

General power function

The function x to the power n for any real n, defined for positive x as e to the power n ln x, so that irrational exponents make sense.

Frequently asked questions

References

1. Joel Hass, Christopher Heil, Maurice D. Weir et al. (2023). Thomas' Calculus: Early Transcendentals in SI Units (15th ed.), Section 3.8: Derivatives of Inverse Functions and Logarithms. Pearson.
2. James Stewart (2015). Calculus (8th ed.), section 3.6: Derivatives of Logarithmic Functions. Cengage Learning.
3. John Napier (1614). Mirifici Logarithmorum Canonis Descriptio. Andrew Hart, Edinburgh.
4. Nicolaus Mercator (1668). Logarithmotechnia. London.
5. Eli Maor (1994). e: The Story of a Number. Princeton University Press.

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