# differentiate question

• Feb 18th 2010, 04:10 AM
decoy808
differentiate question
$\displaystyle y=\sqrt{x^3} (x+1)$

dy/dx = $\displaystyle (x^3){^\frac{1}{3}}+1$

$\displaystyle \frac{1}{3}x^\frac{-1}{2}$

• Feb 18th 2010, 05:38 AM
danielomalmsteen
$\displaystyle y=x^{\frac{2}{3} +1} + x^{\frac{2}{3}}$

then

$\displaystyle y'=\left(\frac{2}{3} +1 \right)\cdot x^{\frac{2}{3} +1 -1} + \left(\frac{2}{3}\right)\cdot x^{\frac{2}{3}-1}$
• Feb 18th 2010, 05:51 AM
NRS
Youre going to have to use the chain rule
First of all, http://www.mathhelpforum.com/math-he...743f6a35-1.gif is not correct, x cubed will be to the one half power, because you are taking the square root of it, not the cube root.

To differentiate, you multiply ((x^3)^(1/2)) and x^(1) to get

(x^3)^(3/2) + 1

Now, differentiate:

dy/dx = (3/2)(x^3)^(1/2)*(3x^2)
• Feb 18th 2010, 06:06 AM
HallsofIvy
You also appear not to have "distributed":

$\displaystyle y= \sqrt{x^3}(x+1)= x^{3/2}(x+1)= x^{5/2}+ x^{3/2}$

Now, use that the derivative of $\displaystyle x^n$ is $\displaystyle nx^{n-1}$.