Question

If $y=(tanx{)}^{(tanx{)}^{tanx}},$ then at $x=\frac{\pi }{4},\frac{dy}{dx}$ is equal to

• A. 0
• B. 3
• C. 2
• D. None of these

Answer 1

The correct option is C 2

$y={\left(tanx\right)}^{{\left(tanx\right)}^{\left(tanx\right)}}$

We have to find $\frac{dy}{dx}$ at $x=\frac{\Pi }{4}$.

Now consider:

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

$ln\left(y\right)=g\left(x\right)lnf\left(x\right)$

$\Rightarrow \frac{1}{y}\frac{dy}{dx}=g\left(x\right)\frac{f^1\left(x\right)}{f\left(x\right)}+g^1\left(x\right)lnf\left(x\right)$

$\Rightarrow \frac{dy}{dx}=y[g\left(x\right)\frac{f^1\left(x\right)}{f\left(x\right)}+g^1\left(x\right)lnf\left(x\right)]$

$\Rightarrow \frac{dy}{dx}={f\left(x\right)}^{g\left(x\right)}[g^1\left(x\right)lnf\left(x\right)+g\left(x\right)\frac{f^1\left(x\right)}{f\left(x\right)}]$

So, We have

$y={\left(tanx\right)}^{{\left(tanx\right)}^{tanx}}$

Where $f\left(x\right)=tan\left(x\right)$

$f^1\left(x\right)={\sec }^{2}x$

$g\left(x\right)={\left(tanx\right)}^{tanx}$

$g^1\left(x\right)={\left(tanx\right)}^{tanx[{\sec }^{2}xln\left(tan\left(x\right)\right)+\frac{\tan \left(x\right){\sec }^{2}x}{tanx}]}$

Now, $\tan \left(\frac{\Pi }{4}\right)=1$ and $\sec \left(\Pi /4\right)=\sqrt{2}$.

$f\left(\Pi /4\right)=1$

$f^1\left(\Pi /4\right)=2$

$g\left(\Pi /4\right)=1$

$g^1\left(\Pi /4\right)=1(2ln\left(1\right)+2)=2$

$\frac{dy}{dx}$ at $\Pi /4$ is

$\frac{dy}{dx}={f\left(\Pi /4\right)}^{g\left(\Pi /4\right)[g^1\left(\Pi /4\right)ln\left(f\left(\Pi /4\right)\right)+g\left(\Pi /4\right)\frac{f^1\left(\Pi /4\right)}{f\left(\Pi /4\right)}]}$

$\Rightarrow \frac{dy}{dx}=1[2ln\left(1\right)+1\times 2]=2$

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