# differentiate question

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

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

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

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

then

$y'=\left(\frac{2}{3} +1 \right)\cdot x^{\frac{2}{3} +1 -1} + \left(\frac{2}{3}\right)\cdot x^{\frac{2}{3}-1}$
• February 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)
• February 18th 2010, 06:06 AM
HallsofIvy
You also appear not to have "distributed":

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

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