Question

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

Answer 1

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

$\Rightarrow ln\left(y\right)=g\left(x\right)ln\left(f\right(x\left)\right)$

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

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

$\Rightarrow \frac{dy}{dx}=f(x{)}^{g\left(x\right)}\left[g^{\prime }\right(x\left)ln\right(f\left(x\right))+g(x)\times \frac{f^{\prime }\left(x\right)}{f\left(x\right)}]$
We have, $y=tan(x{)}^{tan(x{)}^{tan\left(x\right)}}$
where:
$f\left(x\right)=tan\left(x\right)$
$f^{\prime }\left(x\right)=se{c}^2\left(x\right)$
$g\left(x\right)=tan(x{)}^{tan\left(x\right)}$
$g^{\prime }\left(x\right)=tan(x{)}^{tan\left(x\right)}\left[se{c}^2\right(x\left)ln\right(tan\left(x\right))+tan(x)\times \frac{se{c}^2\left(x\right)}{tan\left(x\right)}]$
$=tan\left(x{)}^{tan\left(x\right)}\right[se{c}^2\left(x\right)ln\left(tan\right(x\left)\right)+se{c}^2\left(x\right)]$
Now $tan\left(\pi /4\right)=1$ and $sec\left(\pi /4\right)=\sqrt{2}$
Therefore
$f\left(\pi /4\right)=1$
$f^{\prime }\left(\pi /4\right)=2$
$g\left(\pi /4\right)=1^1=1$
$g^{\prime }\left(\pi /4\right)=1^1\left(2ln\right(1)+2)=2$
And so $\frac{dy}{dx}$ at $\frac{\pi }{4}$ is $dy/dx=f(\pi /4{)}^{g\left(\pi /4\right)}\left[g^{\prime }\right(\pi /4\left)ln\right(f\left(\pi /4\right))+g(\pi /4)\times \frac{f^{\prime }\left(\pi /4\right)}{f\left(\pi /4\right)}]$
$\Rightarrow \frac{dy}{dx}=1^1\left[2ln\right(1)+1\times \frac{2}{1}]$
$\therefore \frac{dy}{dx}=2$