# Find derivative.

• Oct 11th 2010, 07:23 AM
Lil
Find derivative.
$f(x)=(\sqrt{x}-3x^{3})(2\sqrt{x}+5)$.

How I do...
$f'(x)=((\sqrt{x}-3x^{3})(2\sqrt{x}+5))'=$ $(\sqrt{x}-3x^{3})'\cdot (2\sqrt{x}+5)+(\sqrt{x}-3x^{3})\cdot(2\sqrt{x}+5)'=$ $(\frac{1}{2\sqrt{x}}-9x^{2})(2\sqrt{x}+5)+(\sqrt{x}-3x^{3})(\frac{2}{2\sqrt{x}}+0)=$ $\frac{2\sqrt{x}}{2\sqrt{x}}+\frac{5}{2\sqrt{x}}-18x^{2}\sqrt{x}-45x^{2}+\frac{\sqrt{x}}{\sqrt{x}}-\frac{3x^{3}}{\sqrt{x}}=$
$\frac{4\sqrt{x}+5-36x-90x^{2}\sqrt{x}-6x^3}{2\sqrt{x}}$

I don't know it's good or not. This answer looks strange. Maybe I do mistake? Hmm
• Oct 11th 2010, 07:46 AM
HallsofIvy
Almost. $18x^2\sqrt{x}\frac{2\sqrt{x}}{2\sqrt{x}}= \frac{32x^3}{2\sqrt{x}}$, not $\frac{36x}{2
sqrt{x}}$