Question

If $y=tan\hspace{0.17em}{x}^{tan\hspace{0.17em}{x}^{tan\hspace{0.17em}x}}$, then $\frac{dy}{dx}$ is

Answer 1

Step 1: Express using Logarithm

The given expression is,

$y=tan\hspace{0.17em}{x}^{tan\hspace{0.17em}{x}^{tan\hspace{0.17em}x}}$

Taking the logarithm on both sides we get,

$\log y=\tan x^{\tan x}\log \left(\tan x\right)$ $\left[\because \log \left(a^b\right)=b\log a\right]$

Taking the logarithm again on both sides we get,

$\log \left(\log y\right)=\log \left[\tan x^{\tan x}\log \left(\tan x\right)\right]$

$\Rightarrow$$\log \left(\log y\right)=\tan x\log \left(\tan x\right)+\log \left[\log \left(\tan x\right)\right]$ $\left[\because \log \left(a^b\right)=b\log a;\log \left(ab\right)=\log a+\log b\right]$

Step 2: Differentiate with respect to $x$

Upon differentiating both sides with respect to $x$ we get,

$\frac{1}{\log y}.\frac{1}{y}.\frac{dy}{dx}=\tan x.\frac{1}{\tan x}.se{c}^{2}x+\log \left(\tan x\right).se{c}^{2}x+\frac{1}{\log \left(\tan x\right)}.\frac{1}{\tan x}.se{c}^{2}x$ $$\begin{gathered} \left[\because \frac{d}{dx}\left(\log x\right)=\frac{1}{x};\frac{d}{dx}\left(\tan x\right)=se{c}^{2}x;\right] \\ \left[\because \frac{d}{dx}\left(uv\right)=u\frac{d}{dx}\left(v\right)+v\frac{d}{dx}\left(u\right);\right] \\ \left[\because \frac{d}{dx}\left[f\left\{g\left(x\right)\right\}\right]=f\text{'}\left\{g\left(x\right)\right\}.g\text{'}\left(x\right)\right] \end{gathered}$$

$\Rightarrow$$\frac{dy}{dx}.\frac{1}{y\log y}=se{c}^{2}x\left[1+\log \left(\tan x\right)+\frac{1}{\tan x\log \left(\tan x\right)}\right]$

$\Rightarrow$ $\frac{dy}{dx}=y\log yse{c}^{2}x\left[1+\log \left(\tan x\right)+\frac{1}{\tan x\log \left(\tan x\right)}\right]$

Hence, the expression for $\frac{dy}{dx}$ is $y\log yse{c}^{2}x\left[1+\log \left(\tan x\right)+\frac{1}{\tan x\log \left(\tan x\right)}\right]$.