Question

${10}^{-x\tan x}\left[\frac{d}{dx}\left({10}^{x\tan x}\right)\right]$ is equal to

• A. $\tan x–xse{c}^{2}x$
• B. $\log 10[\tan x+xse{c}^{2}x]$
• C. $x\tan x\log 10$
• D. None of these

Answer 1

The correct option is B

$\log 10[\tan x+xse{c}^{2}x]$

Explanation for the correct option:

Step 1. Find the value of ${\mathbf{10}}^{\mathbf{-}\mathbf{x}\mathbf{}\mathbf{t}\mathbf{a}\mathbf{n}\mathbf{}\mathbf{x}}\mathbf{}\left[\frac{d}{dx}\left({10}^{xtanx}\right)\right]$

Let $y={10}^{x\tan x}$

$\Rightarrow$ ${\log }_{10}y=x\tan x$

$\Rightarrow$ $\frac{\log y}{\log 10}=x\tan x$ $\left[\mathbf{\because }\mathit{l}\mathit{o}{\mathit{g}}_{\mathbf{b}}\mathbf{}\mathit{M}\mathbf{}\mathbf{=}\mathbf{}\frac{{\mathbf{log}}_{\mathbf{a}}\mathbf{}\mathbf{M}}{{\mathbf{log}}_{\mathbf{a}}\mathbf{}\mathbf{b}}\right]$

$\Rightarrow$ $\log y=\left(\log 10\right)\left(x\tan x\right)$

Step 2. Differentiate it with respect to ‘$x$’

$\frac{1}{y}\frac{dy}{dx}=\log 10\left[x\left(se{c}^{2}x\right)+\tan x\left(1\right)\right]$ $\left[\mathbf{\because }\frac{\mathbf{d}\left(uv\right)}{\mathbf{d}\mathbf{x}}\mathbf{}\mathbf{=}\mathbf{}\mathit{u}\frac{\mathbf{d}\mathbf{v}}{\mathbf{d}\mathbf{x}}\mathbf{}\mathbf{+}\mathbf{}\mathit{v}\frac{\mathbf{d}\mathbf{u}}{\mathbf{d}\mathbf{x}}\right]$

$\Rightarrow$ $\frac{dy}{dx}=y\times \log 10\left[\tan x+xse{c}^{2}x\right]$

$\Rightarrow$$\frac{d}{dx}\left({10}^{x\tan x}\right)={10}^{x\tan x}\times \log 10\left[\tan x+xse{c}^{2}x\right]$

Step 3. multiply both sides by ${\mathbf{10}}^{\mathbf{-}\mathbf{x}\mathbf{}\mathbf{t}\mathbf{a}\mathbf{n}\mathbf{x}}$

${10}^{-x\tan x}\times \frac{d}{dx}\left({10}^{x\tan x}\right)={10}^{-x\tan x}\times {10}^{x\tan x}\times {\log }_{10}10\left[\tan x+xse{c}^{2}x\right]$ $\mathrm{[}\because a^{-m}\times a^m=1\mathrm{]}$

$\mathbf{\therefore }{\mathbf{10}}^{\mathbf{-}\mathbf{x}\mathbf{}\mathbf{t}\mathbf{a}\mathbf{n}\mathbf{x}}\mathbf{\times }\frac{\mathbf{d}}{\mathbf{d}\mathbf{x}}\mathbf{\left(}{\mathbf{10}}^{\mathbf{x}\mathbf{}\mathbf{t}\mathbf{a}\mathbf{n}\mathbf{}\mathbf{x}}\mathbf{\right)}\mathbf{=}\mathbf{log}\mathbf{10}\mathbf{[}\mathbf{t}\mathbf{a}\mathbf{n}\mathbf{x}\mathbf{+}\mathbf{x}\mathbf{}\mathbf{s}\mathbf{e}{\mathbf{c}}^{\mathbf{2}}\mathbf{x}\mathbf{]}$

Hence, Option ‘B’ is Correct.