Skip to content

Mathematics · Ch 5 — Continuity and Differentiability

Logarithmic Differentiation

5.5

Logarithmic Differentiation

5.5 Logarithmic Differentiation

Why Logarithmic Differentiation?

The standard rules handle xnx^n (variable base, constant exponent) and axa^x (constant base, variable exponent). But for [u(x)]v(x)[u(x)]^{v(x)}, where both base and exponent are functions of xx, neither the power rule nor the exponential rule applies directly. Logarithmic differentiation is the systematic method: take the natural log of both sides of y=f(x)y = f(x), simplify with logarithm properties, then differentiate implicitly.

Important

Prerequisite: f(x)f(x) and the base u(x)u(x) must be strictly positive for all xx considered, or their logarithms are undefined in the real number system.


The General Method

Let y=f(x)=[u(x)]v(x)y = f(x) = [u(x)]^{v(x)}, with u(x)>0u(x) > 0 and v(x)v(x) differentiable.

Step 1 — take natural logarithm (base ee) on both sides:

log⁡y=log⁡([u(x)]v(x))\log y = \log\left([u(x)]^{v(x)}\right)

Step 2 — apply the power rule log⁡(ab)=blog⁡a\log(a^b) = b \log a:

log⁡y=v(x)⋅log⁡[u(x)]\log y = v(x) \cdot \log[u(x)]

Step 3 — differentiate both sides w.r.t. xx (chain rule on the left, product rule on the right):

ddx(log⁡y)=1y⋅dydx\frac{d}{dx}(\log y) = \frac{1}{y} \cdot \frac{dy}{dx}

ddx[v(x)⋅log⁡[u(x)]]=v′(x)⋅log⁡[u(x)]+v(x)⋅u′(x)u(x)\frac{d}{dx}\big[v(x) \cdot \log[u(x)]\big] = v'(x) \cdot \log[u(x)] + v(x) \cdot \frac{u'(x)}{u(x)}

using ddx(log⁡[u(x)])=u′(x)u(x)\frac{d}{dx}\big(\log[u(x)]\big) = \frac{u'(x)}{u(x)} by the chain rule.

Step 4 — equate and multiply both sides by yy:

1y⋅dydx=v′(x)⋅log⁡[u(x)]+v(x)⋅u′(x)u(x)\frac{1}{y} \cdot \frac{dy}{dx} = v'(x) \cdot \log[u(x)] + \frac{v(x) \cdot u'(x)}{u(x)}

dydx=y[v′(x)⋅log⁡[u(x)]+v(x)⋅u′(x)u(x)]\frac{dy}{dx} = y \left[ v'(x) \cdot \log[u(x)] + \frac{v(x) \cdot u'(x)}{u(x)} \right]

Step 5 — substitute back y=[u(x)]v(x)y = [u(x)]^{v(x)}:

dydx=[u(x)]v(x)[v′(x)⋅log⁡[u(x)]+v(x)⋅u′(x)u(x)]\boxed{\frac{dy}{dx} = [u(x)]^{v(x)} \left[ v'(x) \cdot \log[u(x)] + \frac{v(x) \cdot u'(x)}{u(x)} \right]}

Logarithmic Differentiation Formula

ddx([u(x)]v(x))=[u(x)]v(x)[v′(x)log⁡[u(x)]+v(x)u′(x)u(x)]\frac{d}{dx}\big([u(x)]^{v(x)}\big) = [u(x)]^{v(x)} \left[ v'(x) \log[u(x)] + \frac{v(x) u'(x)}{u(x)} \right]


Special Case: Differentiating axa^x (Constant Base)

This is the general formula with u(x)=au(x) = a (constant, a>0a > 0) and v(x)=xv(x) = x.

Method 1 (logarithmic differentiation): Let y=axy = a^x. Then log⁡y=xlog⁡a\log y = x \log a, so …