Question

given the function y = (2 + 4x^2 - 5x) (-3x^3 + 5), find $\frac{dy}{dx}$ in any form.

Explanation:

Step1: Apply product - rule

The product - rule states that if $y = u\cdot v$, where $u = 2 + 4x^{2}-5x$ and $v=-3x^{3}+5$, then $\frac{dy}{dx}=u\frac{dv}{dx}+v\frac{du}{dx}$.
First, find $\frac{du}{dx}$ and $\frac{dv}{dx}$.
$\frac{du}{dx}=\frac{d}{dx}(2 + 4x^{2}-5x)=0 + 8x-5=8x - 5$.
$\frac{dv}{dx}=\frac{d}{dx}(-3x^{3}+5)=-9x^{2}+0=-9x^{2}$.

Step2: Substitute into product - rule formula

$\frac{dy}{dx}=(2 + 4x^{2}-5x)(-9x^{2})+(-3x^{3}+5)(8x - 5)$.
Expand the two products:
$(2 + 4x^{2}-5x)(-9x^{2})=-18x^{2}-36x^{4}+45x^{3}$.
$(-3x^{3}+5)(8x - 5)=-24x^{4}+15x^{3}+40x - 25$.

Step3: Combine like - terms

$\frac{dy}{dx}=(-36x^{4}-24x^{4})+(45x^{3}+15x^{3})-18x^{2}+40x - 25$.
$\frac{dy}{dx}=-60x^{4}+60x^{3}-18x^{2}+40x - 25$.

Answer:

$\frac{dy}{dx}=-60x^{4}+60x^{3}-18x^{2}+40x - 25$