Answer
\begin{aligned}
\frac{d}{dx}\left(\frac{2}{15 b^{2}}(3 b x-2 a)(a+b x)^{3 / 2}+C\right) = x\sqrt{a+b x}
\end{aligned}
Work Step by Step
Given
$$\int x \sqrt{a+b x} d x=\frac{2}{15 b^{2}}(3 b x-2 a)(a+b x)^{3 / 2}+C$$
Since
\begin{aligned}
\frac{d}{dx}\left(\frac{2}{15 b^{2}}(3 b x-2 a)(a+b x)^{3 / 2}+C\right)&=\frac{2}{15 b^{2}}\frac{d}{dx}\left((3 b x-2 a)(a+b x)^{3 / 2}\right)+\frac{d}{dx}(C)\\
&= \frac{2}{15 b^{2}} \left[ (3 b x-2 a)\frac{d}{dx}(a+b x)^{3 / 2}+(a+b x)^{3 / 2}\frac{d}{dx}(3 b x-2 a)\right]+0\\
&= \frac{2}{15 b^{2}} \left[ (3 b x-2 a)(\frac{3}{2}(b)(a+b x)^{1/ 2})+(a+b x)^{3 / 2} (3 b )\right]\\
&= \frac{2}{5 b} \left[ \frac{1}{2} (3 b x-2 a)\sqrt{a+b x}+(a+bx) \sqrt{a+b x} \right]\\
&= \frac{2\sqrt{a+b x}}{5 b} \left[ \frac{3b}{2}x -a+a+bx\right]\\
&= \frac{2\sqrt{a+b x}}{5 b} \left[ \frac{5b}{2}x \right]\\
&= x\sqrt{a+b x}
\end{aligned}
