Question

If $y=e^{x^{e^x}}+x^{e^{e^x}}+e^{x^{x^e}},$ find $\frac{dy}{dx}$

• A. $e^{x^{e^x}}.x^x[\frac{e^x}{x}+e^x\log x]+x^{e^{e^x}}.e^{e^x}[\frac{1}{x}+e^x\log x]+e^{x^e}.x^{x^e}.x^e(1+e\log x)$
• B. $e^{x^{e^x}}.x^{e^x}[\frac{e^x}{x}+e^x\log x]+x^{e^{e^x}}.e^{e^x}[\frac{1}{x}+e^x\log x]+e^{x^{x^e}}.x^{x^e}.x^{e-1}(1+e\log x)$
• C. $x^{e^x}.x^{e^x}[\frac{e^x}{x}+e^x\log x]+e^{e^x}.e^{e^x}[\frac{1}{x}+e^x\log x]+x^{x^e}.x^{x^e}.x^{e-1}(1+e\log x)$
• D. none of these

Answer 1

The correct option is B $e^{x^{e^x}}.x^{e^x}[\frac{e^x}{x}+e^x\log x]+x^{e^{e^x}}.e^{e^x}[\frac{1}{x}+e^x\log x]+e^{x^{x^e}}.x^{x^e}.x^{e-1}(1+e\log x)$

Let $y=e^{x^{e^x}}+x^{e^{e^x}}+e^{x^{x^e}}=u+v+w$
$\Rightarrow \frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}+\frac{dw}{dx}$
$u=e^{x^{e^x}}\Rightarrow \log u=x^{e^x}\therefore \log \left(\log u\right)=e^x\log x.$
Differentiate.$\frac{1}{\log u}.\frac{1}{u}\frac{du}{dx}=e^x\log x+\frac{e^x}{x}$
$\therefore \frac{du}{dx}=u\log u[\frac{e^x}{x}+e^x\log x]=e^{x^{e^x}}.x^{e^x}[\frac{e^x}{x}+e^x\log x].$
Similarly , $\frac{dv}{dx}=x^{e^{e^x}}.e^{e^x}[\frac{1}{x}+e^x\log x]$
and $\frac{dw}{dx}=e^{x^{x^e}}.x^{x^e}.x^{e-1}(1+e\log x)$
$\therefore \frac{dy}{dx}=e^{x^{e^x}}.x^{e^x}[\frac{e^x}{x}+e^x\log x]+x^{e^{e^x}}.e^{e^x}[\frac{1}{x}+e^x\log x]+e^{x^{x^e}}.x^{x^e}.x^{e-1}(1+e\log x)$