Question

differentiate. f(x)=e^{x^{4}} + 9x f(x)=□

Explanation:

Step1: Use sum - rule of differentiation

If $y = u + v$, then $y'=u'+v'$. Here $u = e^{x^{4}}$ and $v = 9x$. So $f'(x)=(e^{x^{4}})'+(9x)'$.

Step2: Differentiate $e^{x^{4}}$ using chain - rule

The chain - rule states that if $y = e^{u}$ and $u = x^{4}$, then $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$. Since $\frac{d}{du}(e^{u}) = e^{u}$ and $\frac{d}{dx}(x^{4})=4x^{3}$, we have $(e^{x^{4}})'=e^{x^{4}}\cdot4x^{3}$.

Step3: Differentiate $9x$

Using the power - rule $\frac{d}{dx}(ax)=a$ (where $a = 9$), we get $(9x)' = 9$.

Step4: Combine the results

$f'(x)=e^{x^{4}}\cdot4x^{3}+9$.

Answer:

$4x^{3}e^{x^{4}} + 9$