Question

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

Explanation:

Step1: Apply sum - rule of differentiation

The sum - rule states that if $y = u + v$, then $y^\prime=u^\prime + v^\prime$. Here, $u = e^{x^{4}}$ and $v = 9x$. So, $f^\prime(x)=\frac{d}{dx}(e^{x^{4}})+\frac{d}{dx}(9x)$.

Step2: Differentiate $v = 9x$

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

Step3: Differentiate $u = e^{x^{4}}$ using chain - rule

The chain - rule states that if $y = f(g(x))$, then $y^\prime=f^\prime(g(x))\cdot g^\prime(x)$. Let $g(x)=x^{4}$ and $f(u)=e^{u}$. Then $f^\prime(u)=e^{u}$ and $g^\prime(x) = 4x^{3}$. So, $\frac{d}{dx}(e^{x^{4}})=e^{x^{4}}\cdot4x^{3}=4x^{3}e^{x^{4}}$.

Step4: Combine the results

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

Answer:

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