Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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:

$y=-3/\sqrt[3]{x}$ Where the ^3 squareroot X = cube root.

Can somebody help? With working shown so that i can understand the process

Thank you!!!

share|improve this question
    
Do you know that $ \frac {d}{dx} x^n = n x^{n-1}$? –  Calvin Lin Jan 14 '13 at 19:13
    
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. –  Robert Israel Jan 14 '13 at 19:17
    
@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. –  Calvin Lin Jan 14 '13 at 19:19

4 Answers 4

up vote 1 down vote accepted

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

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

share|improve this answer
    
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 :( –  user58280 Jan 14 '13 at 19:39
    
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. –  user58280 Jan 14 '13 at 19:41
    
The book's answer must be an answer to a different problem, like the derivative of $\dfrac{3}{4x^5}$ –  amWhy Jan 14 '13 at 19:55
    
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 $^{-1/3}$ –  user58280 Jan 14 '13 at 19:56
    
I read the wrong answer, the answer to this question is $x^{4/3}$ How? –  user58280 Jan 14 '13 at 19:58

$$y'=-3x^{-1/3}=-3\cdot \frac{-1}{3}x^{-1/3-1}=x^{-4/3}=\frac{1}{x^{4/3}}$$

share|improve this answer

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.

share|improve this answer

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:

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

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

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.