Question

question given $f(x)=(2x - 3)^2(5x^{2}+2)^3$, find $f(x)$. select the correct answer below: $\frac{-40x^{2}+98}{(2x - 3)(5x^{2}+2)}$ $2(2x - 3)(5x^{2}+2)^2(40x^{2}-45x + 4)$ $\frac{80x^{2}-82}{(2x - 3)(5x^{2}+2)}$ $12(2x - 3)(5x^{2}+2)^2(40x^{2}+45x + 4)$

Explanation:

Step1: Apply product - rule

Let $u=(2x - 3)^2$ and $v=(5x^2+2)^3$. The product - rule states that $(uv)'=u'v + uv'$. First, find $u'$ using the chain - rule. If $u=(2x - 3)^2$, let $t = 2x-3$, then $u=t^2$. So $u' = 2(2x - 3)\times2=4(2x - 3)$. Second, find $v'$ using the chain - rule. If $v=(5x^2+2)^3$, let $s = 5x^2+2$, then $v=s^3$. So $v'=3(5x^2+2)^2\times10x = 30x(5x^2+2)^2$.

Step2: Calculate $f'(x)$

$f'(x)=u'v+uv'=4(2x - 3)(5x^2+2)^3+(2x - 3)^2\times30x(5x^2+2)^2=(2x - 3)(5x^2+2)^2[4(5x^2+2)+30x(2x - 3)]=(2x - 3)(5x^2+2)^2(20x^2+8 + 60x^2-90x)=(2x - 3)(5x^2+2)^2(80x^2-90x + 8)=2(2x - 3)(5x^2+2)^2(40x^2-45x + 4)$

Answer:

$2(2x - 3)(5x^2+2)^2(40x^2-45x + 4)$