Question

How do you differentiate $f\left(x\right)=(x^2+1)(x+2{)}^2(x-3{)}^3$ using the product rule ?

Answer 1

First split off your separate expressions into sub-functions.

Let $y=t\cdot u\cdot v$

where $t=x^2+1,u=(x+2{)}^2$ , and $v=(x-3{)}^3$

Then $\frac{dx}{dt}=2x$.

$\frac{du}{dx}=2(x+2)$.

By the chain rule, $\frac{dv}{dx}=3(x-3{)}^2$.

The product rule for three terms states.

If $y=t\cdot u\cdot v$, and $u$ is a function of $x$.

Then $\frac{dy}{dx}=\frac{dt}{dx}\cdot u\cdot v+\frac{du}{dx}\cdot t\cdot v+\frac{dv}{dx}\cdot t\cdot u$.

So, $\frac{dy}{dx}=2x\cdot (x+2{)}^2\cdot (x-3{)}^3+2(x^2+1)\cdot (x-3{)}^3\cdot (x+2)+3(x^2+1)\cdot (x+2{)}^2$

Which when you go through the painful process of expansion and simplification, yields:

$\frac{dy}{du}=7x^6-30x^5-20x^4+160x^3-15x^2-126x$