Differentiate 4

$\large \dfrac{d}{dx}{e}^{{x}^{{e}}}$

Suppose $x^e = u$

$\large = \dfrac{d}{du}(e^u) \dfrac{d}{dx} (x^e)$

$\large = e^u ex^{e-1}$

Substituting $u = x^e$

$\large = e^{x^e} ex^{e-1}$

$\large = e^{x^{e}+1} x^{e-1}$

Here is a solution. I used the power rule for one of the steps, which is shown as follows:

$\frac{d}{dx}(x^{a})=a•x^{a-1}$

I also used the rule of multiplying exponents to get my final result - everyone knows that rule.

Here's the solution:

We have $\frac{d}{dx}$ $e^{x^{e}}$ .

We start by applying the chain rule:

$\frac{df(u)}{dx}$ = $\frac{df}{du}$ $•$ $\frac{du}{dx}$

Let $x^{e}=u$ :

= $\frac{d}{du}$ $(e^u)$ $\frac{d}{dx}$ $(x^e)$

Apply the common derivative to the first part:

$\frac{d}{du}$ $(e^u)$ = $e^u$

Apply the power rule to the second part:

$\frac{d}{dx}$ $(x^e)$ = $ex^{e-1}$

Now we have:

= $e^u$ $ex^{e-1}$

Substitute the original value of $u$ back into $u$ .

= $e^{x^{e}}$ $ex^{e-1}$

Apply the exponent rule and simplify:

= $e^{x^{e}+1}$ $x^{e-1}$

Hope this helped any who were confused about this problem!