Find derivative of $\sqrt[3]{\frac{x^2}{x+1}}$ I should find the derivative of $$ \sqrt[3]{\frac{x^2}{x+1}} $$
I know how to deal with general derivatives (patterns for fractions or composition) but I don't know (and it wasn't on my exercises) how to deal with derivative of for example
$$ (x+1)^{1/3} $$
I saw on that forum that in some cases people use inequalities but I think that there is a more general approach...
 A: This is a typical case for using the logarithmic derivative. Denoting this function as $f(x)$, we have:
$$\frac{f'(x)}{f(x)}=\frac13\Bigl(2\cdot\frac 1x-\frac1{x+1}\Bigr)=\frac 13\frac{x+2}{x(x+1)}, $$
and therefore
$$f'(x)=\frac 13\frac{x+2}{x(x+1)}\biggl(\frac{x^2}{x+1}\biggr)^{\mkern-6mu\frac 13} = \frac{x+2}{3(x+1)\sqrt[3]{x(x+1)}} $$
A: Write the whole thing exponentially and then use the derivative of a product + chain rule:
$$\left(\sqrt[3]\frac{x^2}{x+1}\right)'=\left(x^{2/3}(x+1)^{-1/3}\right)'=\frac23x^{-1/3}(x+1)^{-1/3}+\left(-\frac13\right)(x+1)^{-4/3}x^{2/3}=$$
$$=\frac13\left(\frac2{\sqrt[3]{x(x+1)}}-\sqrt[3]{\frac{x^2}{(x+1)^4}}\right)=...etc.$$
The etc. above means you can write many equivalent expressions.
A: Overkill? Implicit differentiation:
$y^3=\frac{x^2}{x+1}.$
Differentiate both sides with respect to $x$:
$3y^2(dy/dx)= \dfrac{2x(x+1)- x^2}{(x+1)^2};$
$3y^2(dy/dx)=\dfrac{x^2+2x}{(x+1)^2};$
$y \not =0:$
$dy/dx=\dfrac{1}{3y^2}\dfrac{x(x+2)}{(x+1)^2};$
Left to do : Epress $y^2$ in terms of $x$.
A: Hint : Use the quotient rule and the chain rule
