Question

differentiate.
f(x)=(7x^2 - 9x)e^x
f(x)=

Explanation:

Step1: Apply product - rule

The product - rule states that if $y = u\cdot v$, then $y^\prime=u^\prime v + uv^\prime$. Let $u = 7x^{2}-9x$ and $v = e^{x}$.

Step2: Differentiate $u$

Differentiate $u = 7x^{2}-9x$ with respect to $x$. Using the power - rule $\frac{d}{dx}(ax^{n})=nax^{n - 1}$, we get $u^\prime=\frac{d}{dx}(7x^{2}-9x)=14x - 9$.

Step3: Differentiate $v$

Differentiate $v = e^{x}$ with respect to $x$. Since $\frac{d}{dx}(e^{x})=e^{x}$, we have $v^\prime = e^{x}$.

Step4: Calculate $f^\prime(x)$

Substitute $u$, $u^\prime$, $v$, and $v^\prime$ into the product - rule formula: $f^\prime(x)=u^\prime v+uv^\prime=(14x - 9)e^{x}+(7x^{2}-9x)e^{x}$.
Factor out $e^{x}$: $f^\prime(x)=e^{x}[(14x - 9)+(7x^{2}-9x)]=e^{x}(7x^{2}+5x - 9)$.

Answer:

$e^{x}(7x^{2}+5x - 9)$