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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

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  • $\begingroup$ Do you know that $ \frac {d}{dx} x^n = n x^{n-1}$? $\endgroup$ Commented Jan 14, 2013 at 19:13
  • $\begingroup$ One normally takes the derivative of a function or an expression, not an equation. If you want to find the derivative of one or both sides of the equation, use the rules for constant multiples and powers. $\endgroup$ Commented Jan 14, 2013 at 19:17
  • $\begingroup$ @user58280 Set $n=-3$ for the LHS. Set $n= \frac {1}{2}$ for the RHS. The main question being, what is your understanding of differentiation? Are you trying to do it from first principles, or are you trying to differentiate from a formula. $\endgroup$ Commented Jan 14, 2013 at 19:19

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$$y \,=\, \dfrac{-3}{\sqrt[\large 3]{x}} \,= \,-3x^{-1/3}$$

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 $$y \,=\,-3x^{-1/3}\;\;\implies\;\;\frac{dy}{dx} = -(1/3)(-3)\,\large x^{-\large\frac13 - 1} = x^{\large -\frac13 - \large\frac33} = x^{-\frac 43} = \frac{1}{x^{(4/3)}}$$

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  • $\begingroup$ This is all very frustrating, i haven't seen that equation before im afraid. I think the work you're portraying is at a higher level to what i am studying. I have just started extended differentiation after learning how to solve indices questions such as $t^a/t^{a+b}$ However i have not been taught that method yet, i'm sorry you're going to have to dumb it down :( $\endgroup$ Commented Jan 14, 2013 at 19:39
  • $\begingroup$ The answer for this question, it says in the book, is $- 15/4x^6$ However i don't know the method of getting to that answer. $\endgroup$ Commented Jan 14, 2013 at 19:41
  • $\begingroup$ The book's answer must be an answer to a different problem, like the derivative of $\dfrac{3}{4x^5}$ $\endgroup$ Commented Jan 14, 2013 at 19:55
  • $\begingroup$ I i understand how you changed the equation to $-3x^{-1/3}$ but after that how do i differentiate with a fraction? i know that when you differentiate, for example, $y = x^2$ it becomes $y = 2x$ or if you differentiate $y = x^{-1}$ it becomes $y = -x^{-2}$, but i have not yet learned how to differentiage $^{-1/3}$ $\endgroup$ Commented Jan 14, 2013 at 19:56
  • $\begingroup$ I read the wrong answer, the answer to this question is $x^{4/3}$ How? $\endgroup$ Commented Jan 14, 2013 at 19:58
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$$y'=-3x^{-1/3}=-3\cdot \frac{-1}{3}x^{-1/3-1}=x^{-4/3}=\frac{1}{x^{4/3}}$$

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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.

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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:

$\displaystyle3y^2\cdot\frac{dy}{dx}=27x^{-2}$

$\displaystyle\Rightarrow \frac{dy}{dx}=9\frac{x^{-2}}{y^2}=9\frac{x^{-2}}{\left(\frac{-3}{x^{1/3}}\right)^2}$

$\displaystyle=\frac{9x^{-2}}{\frac{9}{x^{2/3}}}=x^{-2+2/3}=x^{-4/3}=\frac{3}{\sqrt[3]{x^4}}$

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