Differentiation Using Logarithm - Practice Questions & MCQ

Differentiation Using Logarithm

1. Till now we have studied how to find derivatives of functions of the form $y=(g(x))^n$ (where $n$ is constant), as well as functions of the form $y=b^{g(x)}, b>0$ and $b \neq 1$ (where $b$ is a constant). But how to find the derivatives of functions of the form $y=[f(x)]^{g(x)}$ such as $y=x^x$ or $y=(\tan x)^{\sin (x)}$. These functions require a technique called logarithmic differentiation, which allows us to differentiate any function of the form $h(x)=[f(x)]^{g(x)}$

Let, $\quad \mathrm{y}=(\mathrm{f}(\mathrm{x}))^{\mathrm{g}(\mathrm{x})}$
Take $\log$ both side

$
\log y=g(x) \log f(x)
$
Differentiate concerning x

$
\begin{array}{rlrl}
\frac{1}{y} \cdot \frac{d y}{d x} & =g(x) \cdot \frac{1}{f(x)} \cdot \frac{d}{d x}(f(x))+\log f(x) \cdot \frac{d}{d x}(g(x)) \\
& \therefore \quad \frac{d y}{d x} & =y\left[\frac{g(x)}{f(x)} \cdot \frac{d}{d x}(f(x))+\log f(x) \cdot \frac{d}{d x}(g(x))\right] \\
& \text { or } \quad \frac{d y}{d x} & =(f(x))^{g(x)}\left[\frac{g(x)}{f(x)} \cdot \frac{d}{d x}(f(x))+\log f(x) \cdot \frac{d}{d x}(g(x))\right]
\end{array}
$

Note:

Shortcut to differentiate such functions

$\frac{d y}{d x}=$ Differential of $y$ treating $f(x)$ as constant + Differential of y treating $\mathrm{g}(\mathrm{x})$ as constant.
2. We can use logarithmic differentiation in cases where $y$ is made up of several factors in the numerator and denominator. Logarithm reduces the calculation time and effort in such cases.

Let, $\quad \mathrm{y}=\frac{\mathrm{f}_1(\mathrm{x}) \cdot \mathrm{f}_2(\mathrm{x}) \cdot \mathrm{f}_3(\mathrm{x}) \ldots}{\mathrm{g}_1(\mathrm{x}) \cdot \mathrm{g}_2(\mathrm{x}) \cdot \mathrm{g}_3(\mathrm{x}) \ldots}$
Take $\log$ both side

$
\begin{aligned}
\log y= & {\left[\log _e\left(f_1(x)\right)+\log _e\left(f_2(x)\right)+\log _e\left(f_3(x)\right)+\ldots\right] } \\
& \quad-\left[\log _e\left(g_1(x)\right)+\log _e\left(g_2(x)\right)+\log _e\left(g_3(x)\right)+\ldots\right]
\end{aligned}
$
Differentiating, w.r.t. $x$, we get

$
\begin{aligned}
\frac{1}{\mathrm{y}} \frac{\mathrm{dy}}{\mathrm{dx}}=\left[\frac{\left(\mathrm{f}_1(\mathrm{x})\right)^{\prime}}{\mathrm{f}_1(\mathrm{x})}+\frac{\left(\mathrm{f}_2(\mathrm{x})\right)^{\prime}}{\mathrm{f}_2(\mathrm{x})}+\frac{\left(\mathrm{f}_3(\mathrm{x})\right)^{\prime}}{\mathrm{f}_3(\mathrm{x})}+\ldots\right] \\
-\left[\frac{\left(\mathrm{g}_1(\mathrm{x})\right)^{\prime}}{\mathrm{g}_1(\mathrm{x})}+\frac{\left(\mathrm{g}_2(\mathrm{x})\right)^{\prime}}{\mathrm{g}_2(\mathrm{x})}+\frac{\left(\mathrm{g}_3(\mathrm{x})\right)^{\prime}}{\mathrm{g}_3(\mathrm{x})}+\ldots\right]
\end{aligned}
$