Question

If $y=a^{a^x}$, then $\frac{dy}{dx}=$

• A. $y.a^x{\left(\log a\right)}^2$
• B. $y.a^x.\log a$
• C. ${(y.a^x)}^2$
• D. $(y.a^x)$

Answer 1

The correct option is B $y.a^x.\log a$

$y=a^{x^y}$

taking logarithm on both sides

$\log y=x^y\log a$

again we take logarithm on both sides

$\log \log y=y\left(\log x\right)+\log \log a$

taking derivative on both sides w.r. t $x$

$\frac{1}{\log y}.\frac{1}{y}.\frac{dy}{dx}=\frac{dy}{dx}\log x+y.\frac{1}{x}+0$

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

$\Rightarrow \frac{dy}{dx}=\frac{y^2\log y}{x(1-y\log x.\log y)}$