Question

if (f(x)=(x^{2}-5x + 8)(3x^{2}-2)), find (f(x)).

Explanation:

Step1: Apply product - rule

The product - rule states that if $y = u(x)v(x)$, then $y'=u'(x)v(x)+u(x)v'(x)$. Let $u(x)=x^{2}-5x + 8$ and $v(x)=3x^{2}-2$.

Step2: Find $u'(x)$ and $v'(x)$

$u'(x)=\frac{d}{dx}(x^{2}-5x + 8)=2x-5$, $v'(x)=\frac{d}{dx}(3x^{2}-2)=6x$.

Step3: Calculate $f'(x)$

$f'(x)=(2x - 5)(3x^{2}-2)+(x^{2}-5x + 8)\times6x$.
Expand:
\[

$$\begin{align*} f'(x)&=(2x\times3x^{2}-2x\times2-5\times3x^{2}+5\times2)+(6x^{3}-30x^{2}+48x)\\ &=(6x^{3}-4x - 15x^{2}+10)+(6x^{3}-30x^{2}+48x)\\ &=6x^{3}-4x - 15x^{2}+10+6x^{3}-30x^{2}+48x\\ &=12x^{3}-45x^{2}+44x + 10 \end{align*}$$

\]

Answer:

$f'(x)=12x^{3}-45x^{2}+44x + 10$