Mathematics - Differentiation Question with Solution | TestHub

Let $y=(a^x{)}^y$. Taking the logarithm on both sides, we get:

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

$$\log y=yx\log a$$

Now, taking the logarithm again on both sides:

$$\log (\log y)=\log (yx\log a)$$

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

Differentiating with respect to $x$:

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

Multiplying by $xy\log y$:

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

Rearranging the terms to solve for $\frac{dy}{dx}$:

$$x\frac{dy}{dx}-x\log y\frac{dy}{dx}=y\log y$$

$$\frac{dy}{dx}(x-x\log y)=y\log y$$

$$\frac{dy}{dx}=\frac{y\log y}{x(1-\log y)}$$