$\displaystyle

y=\sqrt{x^3} (x+1)

$

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

$\displaystyle

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

$

im unsure about this can anyone help or correct me? many thanks

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- Feb 18th 2010, 04:10 AMdecoy808differentiate 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}

$

im unsure about this can anyone help or correct me? many thanks - Feb 18th 2010, 05:38 AMdanielomalmsteen
$\displaystyle y=x^{\frac{2}{3} +1} + x^{\frac{2}{3}}$

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 AMNRSYoure 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 AMHallsofIvy
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 $\displaystyle nx^{n-1}$.