Question

let $h(x)=x^{3}+7e^{x}$. $h(x)=square$

Explanation:

Step1: Apply sum - rule of differentiation

The sum - rule states that if $h(x)=u(x)+v(x)$, then $h'(x)=u'(x)+v'(x)$. Here, $u(x)=x^{3}$ and $v(x)=7e^{x}$.

Step2: Differentiate $u(x)=x^{3}$

Using the power - rule $\frac{d}{dx}(x^{n})=nx^{n - 1}$, for $n = 3$, we have $u'(x)=3x^{2}$.

Step3: Differentiate $v(x)=7e^{x}$

The derivative of $e^{x}$ is $e^{x}$, and by the constant - multiple rule $\frac{d}{dx}(cf(x))=c\frac{d}{dx}(f(x))$ where $c = 7$ and $f(x)=e^{x}$, so $v'(x)=7e^{x}$.

Step4: Find $h'(x)$

By the sum - rule, $h'(x)=u'(x)+v'(x)=3x^{2}+7e^{x}$.

Answer:

$3x^{2}+7e^{x}$