Short and Sweet Calculus

3.10 Logarithmic Differentiation

To differentiate a function \(y=f(x)\) with respect to \(x\) and \(f\) composed of products, quotients, and exponents, it is sometimes easier to:

  1. Take the natural logarithm of both sides of \(y=f(x)\).

  2. Expand the right side using the properties of logarithms; specifically \[\ln(ab)=\ln a+\ln b,\quad\ln\left(\frac{a}{b}\right)=\ln a-\ln b,\quad\ln(a^{r})=r\ln a,\] provided that the right-hand sides make sense (i.e. the input of each logarithm is positive).

  3. Take derivative of both sides.

  4. Isolate \(dy/dx\) and replace \(y\) with the original function.

Example 3.23. Find \(y’\) given \[y=\frac{x\sin x}{(2-x)\sqrt{1+x^{4}}}.\]


This would be really messy to differentiate \(y\) directly. Instead we can use logarithmic differentiation.

First we take the natural logarithm of both sides and exapnd the right hand side using the properties of logarithms \[\begin{aligned} \ln y & =\ln\frac{x\sin x}{(2-x)\sqrt{1+x^{4}}}\\ & =\ln x+\ln\sin x-\ln(2-x)-\frac{1}{2}\ln(1+x^{4}).\end{aligned}\] Now we can differentiate both sides: \[\frac{1}{y}y’=\frac{1}{x}+\frac{1}{\sin x}\left(\frac{d}{dx}\sin x\right)-\frac{1}{2-x}\left(\frac{d}{dx}(2-x)\right)-\frac{1}{2(1+x^{4})}\frac{d}{dx}(1+x^{4})\] or \[\frac{y’}{y}=\frac{1}{x}+\frac{\cos x}{\sin x}-\frac{(-1)}{2-x}-\frac{4x^{3}}{2(1+x^{4})}\] Multiplying both sides by \(y\): \[\begin{aligned} y’ & =y\left[\frac{1}{x}+\frac{\cos x}{\sin x}+\frac{1}{2-x}-\frac{2x^{3}}{1+x^{4}}\right]\\ & =\frac{x\sin x}{(2-x)\sqrt{1+x^{4}}}\left[\frac{1}{x}+\frac{\cos x}{\sin x}+\frac{1}{2-x}-\frac{2x^{3}}{1+x^{4}}\right].\end{aligned}\] We have found the solution, but we may want to simplify it further: \[\begin{aligned} y’ & =\frac{\sin x}{(2-x)\sqrt{1+x^{4}}}+\frac{x\cos x}{(2-x)\sqrt{1+x^{4}}}+\frac{x\sin x}{(2-x)^{2}\sqrt{1+x^{4}}}-\frac{2x^{4}\sin x}{(2-x)\sqrt{(1+x^{4})^{3}}}\\ & =\frac{\sin x+x\cos x}{(2-x)\sqrt{1+x^{4}}}+\frac{x\sin x}{(2-x)^{2}\sqrt{1+x^{4}}}-\frac{2x^{4}\sin x}{(2-x)\sqrt{(1+x^{4})^{3}}}.\end{aligned}\]

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