Question

$y={\left(\tan x\right)}^{{\left(\tan x\right)}^{\left(\tan x\right)}}$ , then at $x=\frac{\mathrm{\pi }}{4}$ , the value of $\frac{dy}{dx}=$

• A. $0$
• B. $1$
• C. $2$
• D. None of these

Answer 1

The correct option is C

$2$

Explanation for the correct option:

Step $1$: Finding the first derivative

The given equation is,

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

Take $\log$ on both sides

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

$\Rightarrow$ $\log \left(y\right)={\tan }^{2}x\log \left(\tan x\right)$

Differentiate both sides with respect to $x$, we get

$\frac{d}{dx}\log \left(y\right)=\frac{d}{dx}\left[{\tan }^{2}x\log \left(\tan x\right)\right]$ $\left[\because \frac{d}{dx}\left(f\left(x\right)g\left(x\right)\right)=f\text{'}\left(x\right)g\left(x\right)+f\left(x\right)g\text{'}\left(x\right)\right]$

$\Rightarrow$ $\frac{1}{y}\frac{dy}{dx}=2\tan x{\sec }^{2}x\log \left(\tan x\right)+{\tan }^{2}x\left(\frac{1}{\tan x}\right){\sec }^{2}x$ $\left[\because \frac{d}{dx}\left({\tan }^{2}x\right)=2\tan x{\sec }^{2}x,\frac{d}{dx}\log \left(\tan x\right)=\frac{1}{\tan x}{\sec }^{2}x\right]$

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

$\Rightarrow$ $\frac{dy}{dx}=y\left(2\tan x{\sec }^{2}x\log \left(\tan x\right)+\tan x{\sec }^{2}x\right)$

Step $2$: Finding the value of $\frac{dy}{dx}$ at $x=\frac{\mathrm{\pi }}{4}$

Substitute $x=\frac{\mathrm{\pi }}{4}$ in the above differential equation.

$\frac{dy}{dx}=y\left(2\tan \frac{\mathrm{\pi }}{4}{\sec }^2\frac{\mathrm{\pi }}{4}\log \left(\tan \frac{\mathrm{\pi }}{4}\right)+\tan \frac{\mathrm{\pi }}{4}{\sec }^2\frac{\mathrm{\pi }}{4}\right)$

$\Rightarrow$ $\frac{dy}{dx}=y\left(2\left(1\right){\left(\sqrt{2}\right)}^2\log \left(1\right)+\left(1\right){\left(\sqrt{2}\right)}^2\right)$ $\left[\because \tan \frac{\mathrm{\pi }}{4}=1,\sec \frac{\mathrm{\pi }}{4}=\sqrt{2}\right]$

$\Rightarrow$ $\frac{dy}{dx}=y\left(2\left(1\right)\left(2\right)\left(0\right)+\left(1\right)\left(2\right)\right)$ $\left[\because \log \left(1\right)=0\right]$

$\Rightarrow$ $\frac{dy}{dx}=y\left(0+2\right)$

$\Rightarrow$ $\frac{dy}{dx}=2y$ $.......\left(1\right)$

At $x=\frac{\mathrm{\pi }}{4}$, the value of $y$ is,

$y={\left(\tan \frac{\mathrm{\pi }}{4}\right)}^{{\left(\tan \frac{\mathrm{\pi }}{4}\right)}^{\left(\tan \frac{\mathrm{\pi }}{4}\right)}}$

$\Rightarrow$ $y={\left(1\right)}^{{\left(1\right)}^{\left(1\right)}}$ $\left[\because \tan \frac{\mathrm{\pi }}{4}=1\right]$

$\Rightarrow$ $y=1$

So, from the equation $\left(1\right)$, we get

$\frac{dy}{dx}=2\left(1\right)$

$\Rightarrow$ $\frac{dy}{dx}=2$

Hence, the correct option is $C)2$.