Derivatives of logarithms and logarithmic differentiation

The natural logarithm, by implicit differentiation. Let y = \ln x for x > 0, so that e^y = x, and take for the moment that \ln is differentiable (proved in the next paragraph). Differentiate both sides with respect to x. On the left, the chain rule treats y as a function of x, exactly as in Implicit differentiation:

The last step replaces e^y by x, which is the equation itself.

A check from the limit definition, which also proves differentiability. Fix x > 0 and take h \ne 0 small enough that x + h > 0. The quotient law gives

Put \displaystyle k = \ln\left(1 + \frac{h}{x}\right). Then \displaystyle 1 + \frac{h}{x} = e^k, so h = x\left(e^k - 1\right), and the quotient becomes

As h \to 0, k \to 0, because \ln is continuous at 1; and k \ne 0 whenever h \ne 0, because \ln is one-to-one. The quotient \displaystyle \frac{e^k - 1}{k} tends to 1, the slope of e^x at 0 (Derivatives of exponential functions). So the difference quotient tends to \displaystyle \frac{1}{x}: the logarithm is differentiable, and its derivative is \displaystyle \frac{1}{x}, as the implicit argument found.

With an absolute value. For x > 0, \ln|x| = \ln x. For x < 0, \ln|x| = \ln(-x), and the chain rule gives \displaystyle \frac{1}{-x} \cdot (-1) = \frac{1}{x}.

Other bases. \displaystyle \log_a x = \frac{1}{\ln a} \cdot \ln x with \ln a a nonzero constant, so its derivative is \displaystyle \frac{1}{x \ln a}. Running the same idea on a^x = e^{x \ln a} gives \big(a^x\big)' = a^x \ln a, already derived in Derivatives of exponential functions and The chain rule.

A logarithm of a function. The outer function is \ln|u|, with derivative \displaystyle \frac{1}{u}, and the chain rule multiplies by u' = g'(x).

Logarithmic differentiation. If f is differentiable at x and f(x) \ne 0, the previous result gives \displaystyle \frac{d}{dx}\ln|f(x)| = \frac{f'(x)}{f(x)}; multiply by f(x). The method assumes f is differentiable. For a variable power this is guaranteed by writing u^v = e^{v \ln u}, which holds because e^{\ln u} = u; it is a composition of differentiable functions, and the chain and product rules give \displaystyle e^{v \ln u}\left(v' \ln u + v \cdot \frac{u'}{u}\right). Logarithmic differentiation of y = u^v, from \ln y = v \ln u, gives the same.

What is stated, not proved. That \ln is continuous, which holds because it is the inverse of the continuous one-to-one function e^x; the general fact belongs with Derivatives of inverse functions.

1. f'(x) = \dfrac{2}{x} for x \ne 0. \ln(3x^2) = \ln 3 + 2\ln|x|, and \displaystyle \big(\ln|x|\big)' = \frac{1}{x}.
2. g'(x) = \dfrac{2x}{(x^2 + 1)\ln 3}. \displaystyle g(x) = \frac{\ln(x^2 + 1)}{\ln 3}, and the chain rule gives \displaystyle \frac{2x}{x^2 + 1} on top.
3. h'(x) = \ln x. The product rule gives \displaystyle \ln x + x \cdot \frac{1}{x} - 1.
4. p'(t) = \dfrac{1 - 2\ln t}{t^3}. The quotient rule gives \displaystyle \frac{t^2 \cdot \frac{1}{t} - 2t\ln t}{t^4}; divide top and bottom by t.
5. q'(x) = \cot x. \displaystyle \frac{u'}{u} = \frac{\cos x}{\sin x}.
6. r'(x) = \dfrac{2}{2x - 5} for \displaystyle x \ne \frac{5}{2}. With u = 2x - 5, \displaystyle \frac{u'}{u}; the absolute value covers \displaystyle x < \frac{5}{2}.
7. s'(x) = \dfrac{3(\ln x)^2}{x}. The power rule with the chain rule: \displaystyle 3(\ln x)^2 \cdot \frac{1}{x}.
8. \displaystyle f'(x) = \dfrac{1}{\sqrt{x^2 + 1}} for every x (the inside is positive, since \displaystyle \sqrt{x^2 + 1} > |x|). \displaystyle u' = 1 + \frac{x}{\sqrt{x^2 + 1}} = \frac{\sqrt{x^2 + 1} + x}{\sqrt{x^2 + 1}}, and dividing by \displaystyle u = x + \sqrt{x^2 + 1} cancels that factor.
9. y' = \dfrac{8}{x(x^2 + 4)} for x \ne 0. y = 2\ln|x| - \ln(x^2 + 4), so \displaystyle y' = \frac{2}{x} - \frac{2x}{x^2 + 4} = \frac{2(x^2 + 4) - 2x^2}{x(x^2 + 4)}.
10. y = 4x - 8. y(2) = \ln 1 = 0 and \displaystyle y' = \frac{2x}{x^2 - 3} = 4 at x = 2.
11. y = 3 + \dfrac{x - 8}{8 \ln 2}. \log_2 8 = 3, and the slope is \displaystyle \frac{1}{8 \ln 2} \approx 0.180.
12. At (3, \ln 3); the tangent is y = \dfrac{x}{3} - 1 + \ln 3. The slope \displaystyle \frac{1}{x} equals \displaystyle \frac{1}{3} only at x = 3; then \displaystyle y - \ln 3 = \frac{1}{3}(x - 3).
13. \dfrac{dM}{dE} = \dfrac{2}{3 \cdot 10^{13} \ln 10} \approx 2.895 \times 10^{-14} per joule, and the rate is 10 times larger. The rate \displaystyle \frac{2}{3E \ln 10} is inversely proportional to E, so ten times the energy gives a tenth of the rate.
14. y' = \dfrac{(x + 2)^3 (x - 1)^4}{(x + 5)^2}\left(\dfrac{3}{x + 2} + \dfrac{4}{x - 1} - \dfrac{2}{x + 5}\right) and y'(2) = \dfrac{2000}{343}. \displaystyle y(2) = \frac{64}{49} and the bracket is \displaystyle \frac{3}{4} + 4 - \frac{2}{7} = \frac{125}{28}.
15. \displaystyle y' = x^2 \sqrt[3]{x^2 + 4}\, e^{3x}\left(\dfrac{2}{x} + \dfrac{2x}{3(x^2 + 4)} + 3\right). From \displaystyle \ln y = 2\ln x + \frac{1}{3}\ln(x^2 + 4) + 3x.
16. y' = \dfrac{1}{(x + 1)^{3/2}(x - 1)^{1/2}}. \displaystyle \ln y = \frac{1}{2}\big[\ln(x - 1) - \ln(x + 1)\big], so \displaystyle \frac{y'}{y} = \frac{1}{2} \cdot \frac{2}{x^2 - 1} = \frac{1}{x^2 - 1}; multiply by \displaystyle y = \frac{(x - 1)^{1/2}}{(x + 1)^{1/2}}.
17. y' = x^{\sin x}\left(\cos x \ln x + \dfrac{\sin x}{x}\right), and the value is 1. From \ln y = \sin x \ln x; at \displaystyle \frac{\pi}{2}, \displaystyle y = \frac{\pi}{2} and the bracket is \displaystyle 0 + \frac{2}{\pi}.
18. y' = x^{1/x} \cdot \dfrac{1 - \ln x}{x^2}; horizontal at \left(e, e^{1/e}\right) \approx (2.718, 1.445). From \displaystyle \ln y = \frac{\ln x}{x} and the quotient rule; x^{1/x} > 0, so y' = 0 only where \ln x = 1.
19. y' = (\ln x)^x\left(\ln(\ln x) + \dfrac{1}{\ln x}\right) and y'(e) = 1. From \ln y = x\ln(\ln x), whose derivative is \displaystyle \ln(\ln x) + x \cdot \frac{1}{x \ln x}; at e, 1^e(0 + 1) = 1.
20. y'(1) = 2\ln 2 + 2 \approx 3.386. \displaystyle \frac{y'}{y} = \ln(x^2 + 1) + \frac{2x^2}{x^2 + 1}; at x = 1, y = 2 and the bracket is \ln 2 + 1.
21. y = 27 + 27(1 + \ln 3)(x - 3). y(3) = 27 and y'(3) = 27(\ln 3 + 1) \approx 56.66.
22. By the product law, \ln(kx) = \ln k + \ln x, and the constant \ln k has derivative 0. By the chain rule, \displaystyle \frac{1}{kx} \cdot k = \frac{1}{x}. The graphs of y = \ln(kx) are vertical shifts of y = \ln x by \ln k, so at each x they all have the same slope.
23. \ln y = \ln f + \ln g gives \displaystyle \frac{y'}{y} = \frac{f'}{f} + \frac{g'}{g}, and multiplying by y = fg gives f'g + fg'. For the quotient, \displaystyle \frac{y'}{y} = \frac{f'}{f} - \frac{g'}{g}, and multiplying by \displaystyle \frac{f}{g} gives \displaystyle \frac{f'}{g} - \frac{fg'}{g^2} = \frac{f'g - fg'}{g^2}.
24. Differentiating \ln y = r \ln x gives \displaystyle \frac{y'}{y} = \frac{r}{x}, so \displaystyle y' = x^r \cdot \frac{r}{x} = r x^{r - 1}; y is differentiable because x^r = e^{r \ln x}. The earlier rules cover integer powers and, through implicit differentiation, rational ones. No root or integer power defines \displaystyle x^{\sqrt{2}}: it is defined as \displaystyle e^{\sqrt{2}\ln x}, so its derivative must go through the logarithm.
25. The derivative is x^x(\ln x + 1) = x^x + x^x \ln x, and x \cdot x^{x - 1} = x^x, so it is the sum of the two pieces. Holding the exponent fixed at a constant r, the power rule gives \displaystyle \frac{d}{dx}x^r = r x^{r - 1}, which at r = x is x \cdot x^{x - 1}. Holding the base fixed at a constant a, the exponential rule gives \displaystyle \frac{d}{dx}a^x = a^x \ln a, which at a = x is x^x \ln x. In \ln y = x \ln x the product rule produces the same two terms: 1 \cdot \ln x from the moving exponent and \displaystyle x \cdot \frac{1}{x} from the moving base. At x = 2 the pieces are 4 and 4\ln 2 \approx 2.77, and the derivative is 4 + 4\ln 2 \approx 6.77.