Logarithmic Differentiation

MATTechniques of Differentiation

Logarithmic differentiation is used when a function has a variable in both base and exponent (e.g. $x^x$ ) or is a large product/quotient of factors.

By taking natural logs first, products turn into sums and exponents drop down as multipliers, after which we differentiate implicitly. It is essential whenever the power rule and exponential rule cannot be applied directly.

Core technique

y = f(x)^{g(x)} \;\Rightarrow\; \log y = g(x)\log f(x)

Differentiate both sides w.r.t.

x

using the chain rule on

\log y

, giving

\dfrac{1}{y}\dfrac{dy}{dx}

on the left; requires

f(x)>0

Standard result for

x^x

\dfrac{d}{dx}\,x^{x} = x^{x}\,(1+\log x)

x>0

; from

\log y = x\log x

\dfrac1y\dfrac{dy}{dx}=\log x + 1

Power-function derivative

\dfrac{d}{dx}\,(f(x))^{g(x)} = (f(x))^{g(x)}\left[g'(x)\log f(x) + g(x)\dfrac{f'(x)}{f(x)}\right]

General formula (

f(x)>0

); reduces to the power rule when

g

is constant and to the exponential rule when

f

is constant.

Sum of two such terms

y = u + v \;\Rightarrow\; \dfrac{dy}{dx} = \dfrac{du}{dx} + \dfrac{dv}{dx}

For

y=x^x+(\sin x)^x

, differentiate each term separately by logarithmic differentiation, then add.

Use logarithmic differentiation whenever the exponent itself is a function of $x$ (variable-base, variable-exponent).
For a sum like $x^x+(\sin x)^x$ , treat each term as a separate function $u$ and $v$ ; never take $\log$ of a sum.
After differentiating $\log y$ , multiply through by $y$ and substitute back the original expression for $y$ .
Use the natural logarithm ( $\log_e$ ); $\dfrac{d}{dx}\log f(x)=\dfrac{f'(x)}{f(x)}$ .
For $(\cos x)^{\sin x}$ , $\log y=\sin x\,\log\cos x$ , giving $\dfrac{dy}{dx}=(\cos x)^{\sin x}\!\left[\cos x\log\cos x-\sin x\tan x\right]$ .
Long products/quotients (many factors) simplify dramatically once you take logs, converting them to sums of $\log$ terms.
State the domain restriction (base $>0$ ) so that $\log$ of the base is defined.

Where the marks go

Taking the logarithm of a sum, e.g. writing $\log(x^x+(\sin x)^x)$ and trying to split it (invalid).
Treating $x^x$ as $x\cdot x^{x-1}$ (power rule) or as $x^x\log x$ (exponential rule) instead of the correct $x^x(1+\log x)$ .
Forgetting to multiply back by $y$ after computing $\dfrac{1}{y}\dfrac{dy}{dx}$ .
Dropping the chain-rule factor when differentiating $\log f(x)$ , e.g. omitting $f'(x)$ in $\dfrac{f'(x)}{f(x)}$ .

How the board asks it

Numericalthe standard result for $x^x$ (variable base and exponent)
If $y = x^x$ , find $\dfrac{dy}{dx}$ .
Numericala sum of two variable-power terms, each handled separately as $u+v$
Find $\dfrac{dy}{dx}$ if $y = x^{\sin x} + (\sin x)^{\cos x}$ .
Numericala long product/quotient converted to a sum of $\log$ terms
Differentiate $y = \dfrac{(x-1)(x-2)}{\sqrt{(x-3)(x-4)(x-5)}}$ with respect to $x$ using logarithmic differentiation.
Derive / provetaking logs of both sides, then implicit differentiation
If $x^y = e^{x-y}$ , prove that $\dfrac{dy}{dx} = \dfrac{\log x}{(1+\log x)^2}$ .
Numericalimplicit differentiation after taking logs of both sides
If $x^y \cdot y^x = 1$ , find $\dfrac{dy}{dx}$ in terms of $x$ and $y$ .
Numerical $\dfrac{d}{dx}\log f(x)=\dfrac{f'(x)}{f(x)}$ , then substitute a value
If $y = (\log x)^x + x^{\log x}$ , find the value of $\dfrac{dy}{dx}$ at $x = e$ .

Practise this topic

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Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.