Question

c. $\frac{d}{dx}((x + 4)(x + 5)(x^{2}+6))$

Explanation:

Step1: Use product - rule

Let $u=(x + 4)(x + 5)$ and $v=x^{2}+6$. The product - rule states that $\frac{d(uv)}{dx}=u\frac{dv}{dx}+v\frac{du}{dx}$. First, find $\frac{dv}{dx}$ and $\frac{du}{dx}$.
$\frac{dv}{dx}=\frac{d(x^{2}+6)}{dx}=2x$

Step2: Expand $u$

$u=(x + 4)(x + 5)=x^{2}+5x+4x + 20=x^{2}+9x + 20$

Step3: Differentiate $u$

$\frac{du}{dx}=\frac{d(x^{2}+9x + 20)}{dx}=2x+9$

Step4: Apply product - rule

$\frac{d((x + 4)(x + 5)(x^{2}+6))}{dx}=(x^{2}+9x + 20)\times(2x)+(x^{2}+6)\times(2x + 9)$
$=2x^{3}+18x^{2}+40x+2x^{3}+9x^{2}+12x + 54$
$=4x^{3}+27x^{2}+52x + 54$

Answer:

$4x^{3}+27x^{2}+52x + 54$