# 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?

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You've done the hardest part. What are we always told to do when we see a sum of two fractions? – 1015 Feb 19 '13 at 16:08
Common denominator? – Hans Groeffen Feb 19 '13 at 16:09
Yes! Precisely. – 1015 Feb 19 '13 at 16:09

Hint: Use the a common denominator $2\sqrt{1+x}$ to add the two terms.

$$f'(x)=2x\sqrt{1+x}+\dfrac{x^2}{2\sqrt{1+x}}\quad =\quad \dfrac {? \quad+ \quad ?}{2\sqrt{1+x}}$$

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Oh I think I got it! $\dfrac{2\times 2x(1+x)+x^2}{2\times\sqrt{1+x}}$ = $\dfrac{4x+5x^2}{2\sqrt{1+x}} \therefore a=5, b=4$. Is that right? – Hans Groeffen Feb 19 '13 at 16:12
Yup...now simplify to get it to the form in which you can solve for $a$ and $b$ – amWhy Feb 19 '13 at 16:13

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: $$f'(x) \; = \; \frac{1}{2} \left(x^4 + x^5 \right)^{-\frac{1}{2}} \cdot \left(4x^3 + 5x^4\right)$$ $$= \;\; \frac{4x^3 + 5x^4}{2\sqrt{x^4 + x^5}}$$ $$= \;\; \frac{4x^3 + 5x^4}{2x^2\sqrt{1 + x}}$$ $$= \;\; \frac{4x + 5x^2}{2\sqrt{1 + x}}$$ $$= \;\; \frac{x\left(4 + 5x\right)}{2\sqrt{1 + x}}$$

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Hint: factor out the $1/\sqrt{1+x}$ term. The term with $\sqrt{1+x}$ will see this factor become $1+x$.

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Something like this you mean? $\sqrt{1+x}\left(2x+\dfrac{x^2}{2\times{(1+x)}^{\frac{3}{2}}}\right)$ – Hans Groeffen Feb 19 '13 at 16:08
No. Read my post again. – Ron Gordon Feb 19 '13 at 16:15