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Mathematics · Ch 5 — Continuity and Differentiability

Logarithmic Differentiation

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Logarithmic Differentiation

The problem this technique solves. None of the differentiation rules developed so far

directly handle a function of the form y=[f(x)]g(x)y=[f(x)]^{g(x)}, where both the base f(x)f(x) and the

exponent g(x)g(x) are themselves functions of xx -- for instance, y=xxy=x^x. The power rule

d(xn)/dx=nxn−1d(x^n)/dx=nx^{n-1} requires a constant exponent nn; the exponential rule

d(ax)/dx=axln⁡ad(a^x)/dx=a^x\ln a requires a constant base aa. In y=xxy=x^x, neither is constant, so

neither rule applies as it stands. Logarithmic differentiation resolves exactly this

situation, by using the property ln⁡(fg)=gln⁡f\ln(f^g)=g\ln f (Section 6) to move the variable exponent down

into an ordinary product, which the chain and product rules can then handle.

Method (variable base and exponent). Given y=[f(x)]g(x)y=[f(x)]^{g(x)} with f(x)>0f(x)>0:

  1. Take the natural logarithm of both sides: ln⁡y=g(x) ln⁡(f(x))\ln y = g(x)\,\ln\big(f(x)\big).
  2. Differentiate both sides with respect to xx, using implicit differentiation on the left (since yy is a function of xx: d(ln⁡y)/dx=(1/y) dy/dxd(\ln y)/dx = (1/y)\,dy/dx, by the chain rule of Section 3) and the product rule on the right:

1ydydx=g′(x)ln⁡(f(x))+g(x)⋅f′(x)f(x).\frac{1}{y}\frac{dy}{dx} = g'(x)\ln\big(f(x)\big) + g(x)\cdot\frac{f'(x)}{f(x)}.

  1. Multiply both sides by y=[f(x)]g(x)y=[f(x)]^{g(x)} to isolate dy/dxdy/dx.

Worked illustration (y=xxy=x^x, x>0x>0). Here f(x)=g(x)=xf(x)=g(x)=x. Taking logs: ln⁡y=xln⁡x\ln y=x\ln x.

Differentiating (product rule on the right, since both factors depend on xx):

1ydydx=1⋅ln⁡x+x⋅1x=ln⁡x+1.\frac{1}{y}\frac{dy}{dx} = 1\cdot\ln x + x\cdot\frac{1}{x} = \ln x+1.

Multiplying back by y=xxy=x^x:

dydx=xx(1+ln⁡x),\frac{dy}{dx} = x^x\big(1+\ln x\big),

matching Example 7.

A second, equally important use: products and quotients of several factors. Even when the

exponents are all constant, logarithmic differentiation turns a lengthy product or quotient --

which would otherwise need repeated applications of the product and quotient rules, with a high

chance of a sign or term-dropping slip -- into a sum of simple logarithmic terms, each

differentiated independently. For y=u1u2⋯umv1v2⋯vny=\dfrac{u_1u_2\cdots u_m}{v_1v_2\cdots v_n} (each ui,vju_i,v_j

a function of xx), taking logs gives

ln⁡y=ln⁡u1+⋯+ln⁡um−ln⁡v1−⋯−ln⁡vn,\ln y = \ln u_1+\cdots+\ln u_m-\ln v_1-\cdots-\ln v_n,

so, differentiating term by term,

1ydydx=u1′u1+⋯+um′um−v1′v1−⋯−vn′vn,hencedydx=y[∑ui′ui−∑vj′vj].\frac{1}{y}\frac{dy}{dx} = \frac{u_1'}{u_1}+\cdots+\frac{u_m'}{u_m} -\frac{v_1'}{v_1}-\cdots-\frac{v_n'}{v_n}, \qquad\text{hence}\qquad \frac{dy}{dx} = y\left[\sum\frac{u_i'}{u_i}-\sum\frac{v_j'}{v_j}\right].

When to reach for logarithmic differentiation. Two clear signals: (i) the function has a

variable raised to a variable power (like xxx^x, xsin⁡xx^{\sin x}, (sin⁡x)x(\sin x)^x), where no ordinary …