How to simplify this derivative.

Question

I am going over a problem in a textbook and I'm having some trouble seeing how this simplifies:

$$f(x) = \frac{x^2}{\sqrt{x+1}}$$

$$f'(x) = \frac{\sqrt{x+1}(2x) - x^2 \cdot \frac{1}{2\sqrt{x+1}}}{x+1}$$

How does this become: $$ = \frac{3x^2 + 4x}{2(x+1)^\frac{3}{2}}$$

Can someone show me how to do the second derivative too? I'm a bit stuck on longer derivations.

Answer 1

$$f'(x) = \frac{\sqrt{x+1}(2x) - x^2 \cdot \frac{1}{2\sqrt{x+1}}}{x+1}=$$ $$=\frac{1}{x+1}\left(\frac{4x(x+1)-x^2}{2\sqrt{x+1}} \right)= $$ $$ =\frac{4x^2+4x-x^2}{2(x+1)\sqrt{x+1}}=\cdots. $$

Answer 2

HINT: multiply numerator and denominaor by $$2\sqrt{x^2+1}$$