If $y=x^{(e^x)}$, then which of the following is an expression for $\frac{dy}{dx}$?

Possible answers could be:

A: $e^x \ln(x)+ e^x/x$

B: $e^x x^{e^x}(\ln(x)+1/x)$

C: $x^{e^x}(e^x \ln(x) - 1/x)$

For my answer I got C but I'm not sure if that's right. Is there anyone who knows the correct answer?


$$y = x^{e^x}$$

$$ y = e^{\log(x)e^x}$$ The chain rule $$\dfrac{dy}{dx} = \dfrac{dy}{du} \dfrac{du}{dx} $$

Let $u = \log(x)e^x$

$$ y = e^u$$

$$ \dfrac{dy}{dx} = e^u \dfrac{du}{dx}$$

$$\dfrac{dy}{dx} = e^u \dfrac{d \,\log(x)e^x}{dx} $$

$$\dfrac{dy}{dx} = x^{e^x} \left( \frac{e^x}{x} + e^x\log(x) \right)$$

$$\dfrac{dy}{dx} = x^{e^x} e^x \left( \frac1{x} + \log(x) \right)$$

Answer is B.


With implicit differentiation ($x > 0$): $$ y(x) = x^{e^x} \quad \Longrightarrow \quad \log y(x) = e^x \log x. $$ Then, $$ \frac{d}{dx} \Big(\log y(x)\Big) = \frac{y'(x)}{y(x)} = \frac{d}{d x}\Big( e^x \log x\Big) = e^x\left(\log x + \frac{1}{x}\right). $$ Hence $$ y'(x) = y(x)e^x\left(\log x + \frac{1}{x}\right) = x^{e^x}e^x\left(\log x + \frac{1}{x}\right). $$


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