$\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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$\displaystyle y=x^{\frac{2}{3} +1} + x^{\frac{2}{3}}$ then $\displaystyle y'=\left(\frac{2}{3} +1 \right)\cdot x^{\frac{2}{3} +1 -1} + \left(\frac{2}{3}\right)\cdot x^{\frac{2}{3}-1}$
First of all, 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)
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 $\displaystyle nx^{n-1}$.
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