Differentiation Of fractions and Negative Powers: $y=-3/\sqrt[3]{x}$ I am stuck with this concept, just can't get my head around it.
I am differentiating fractions and negative powers, I have this question to solve which I just cannot figure out:
$$y=-3/\sqrt[3]{x}$$  
Can somebody help? With working shown so that i can understand the process
 A: $$y \,=\, \dfrac{-3}{\sqrt[\large 3]{x}} \,= \,-3x^{-1/3}$$
Now use the power rule to differentiate y with respect to x: 

Let $c$ be a constant. Then for any $n \neq 0$, (including $n = \pm \dfrac ab $, where $a, b\neq 0$ )
  $$\large y = c\,x^{\,n} \;\implies\; \frac{dy}{dx} =n\cdot c\,x^{\,n-1}$$

In your case, $\;c = -3,\;$ and $\;n\, =\, -\dfrac{1}{3}$.
Applied here, we have $$y \,=\,-3x^{-1/3}\;\;\implies\;\;\frac{dy}{dx} = -(1/3)(-3)\,\large x^{-\large\frac13 - 1} = x^{\large -\frac13 - \large\frac33} = x^{-\frac 43} = \frac{1}{x^{(4/3)}}$$
A: $$y'=-3x^{-1/3}=-3\cdot \frac{-1}{3}x^{-1/3-1}=x^{-4/3}=\frac{1}{x^{4/3}}$$
A: Take the identities d(fg)/dx=d(f)g+fd(g), dx/dx=1, and dc/dx=0 for constants c. From these you can not only derive the formula amWhy gave for integer powers (not zero), but also show it holds for other powers, by taking advantage of the fact that a more elaborate expression like 2x/2 will yield a more elaborate derivative.
A: Just in case anybody hasn't seen how to prove the power rule for fractional powers...
$\displaystyle y(x)=-3\frac{1}{\sqrt[3]{x}}\Rightarrow [y(x)]^3=-27\frac{1}{x}=-27x^{-1}$.
Now differentiate implicitly with respect to $x$ using the Chain Rule:
$\displaystyle3y^2\cdot\frac{dy}{dx}=27x^{-2}$
$\displaystyle\Rightarrow \frac{dy}{dx}=9\frac{x^{-2}}{y^2}=9\frac{x^{-2}}{\left(\frac{-3}{x^{1/3}}\right)^2}$
$\displaystyle=\frac{9x^{-2}}{\frac{9}{x^{2/3}}}=x^{-2+2/3}=x^{-4/3}=\frac{3}{\sqrt[3]{x^4}}$
