Derivative of $x^2\sqrt{1+x}$ 
Given that $f(x)=x^2\sqrt{1+x}$, show that $f'(x)=\dfrac{x(ax+b)}{2\sqrt{1+x}}$ where $a$ and $b$ are constants to be found.

I first tried using the product rule: $f'(x)=2x\sqrt{1+x}+\dfrac{x^2}{2\sqrt{1+x}}$, but now I'm stuck.
I don't know if it is my algebra or derivation that stops me. Could anyone give a hint?

You can avoid the product rule by putting everything under one square root:
$$x^{2} \sqrt{1 + x} \; = \; \sqrt{x^4(1 + x)} \; = \; \sqrt{x^4 + x^5} \; = \; \left(x^4 + x^5 \right)^{\frac{1}{2}}$$
Now apply the power rule for derivatives:
$$f'(x) \; = \; \frac{1}{2} \left(x^4 + x^5 \right)^{-\frac{1}{2}} \cdot \left(4x^3 + 5x^4\right)$$
$$ = \;\; \frac{4x^3 + 5x^4}{2\sqrt{x^4 + x^5}}$$
$$ = \;\; \frac{4x^3 + 5x^4}{2x^2\sqrt{1 + x}}$$
$$ = \;\; \frac{4x + 5x^2}{2\sqrt{1 + x}}$$
$$ = \;\; \frac{x\left(4 + 5x\right)}{2\sqrt{1 + x}}$$
