Answer
$$\frac{d^2}{dx^2}(x^2+9)^5
=10(x^2+9)^4+80x^2(x^2+9)^3.$$
Work Step by Step
Recall that $(x^n)'=nx^{n-1}$
We have
$$\frac{d}{dx}(x^2+9)^5=5(x^2+9)^4(2x)=10x(x^2+9)^4$$
and hence
$\frac{d^2}{dx^2}(x^2+9)^5=10\frac{d}{dx} x(x^2+9)^4$
Use the product rule:
$=10(x^2+9)^4+40x(x^2+9)^3(2x)$
$=10(x^2+9)^4+80x^2(x^2+9)^3$
