if $f(x)=16e^{x}+x^{e}$, find $f(x)$.

Explanation:

Step1: Recall derivative rules

The derivative of $a e^{x}$ is $a e^{x}$ (where $a = 16$ here) and the derivative of $x^{n}$ is $nx^{n - 1}$ (where $n = e$ here).

Step2: Differentiate $16e^{x}$

The derivative of $16e^{x}$ with respect to $x$ is $16e^{x}$ since the derivative of $e^{x}$ is $e^{x}$ and by the constant - multiple rule.

Step3: Differentiate $x^{e}$

Using the power rule $\frac{d}{dx}(x^{n})=nx^{n - 1}$, for $n = e$, the derivative of $x^{e}$ is $ex^{e - 1}$.

Step4: Combine the derivatives

By the sum - rule of differentiation $\frac{d}{dx}(u + v)=\frac{du}{dx}+\frac{dv}{dx}$, where $u = 16e^{x}$ and $v=x^{e}$, we have $f'(x)=16e^{x}+ex^{e - 1}$.

Answer:

$16e^{x}+ex^{e - 1}$