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$f(x)=x^{\frac{1}{3}}$ where $x=27$
i did it in this way
=$3\sqrt{27}$
=3

but the answer is ${\frac{1}{27}}$
this chapter name is (differentiationg rational power $x^{\frac{p}{q}}$)
Can you please help me to solve this question?
thanks in advance

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    First find $f'(x)$: $f'(x)={1\over3}x^{{1\over3}-1}={1\over3} x^{-2/3}={{1\over 3 x^{2/3}}}$. Now evaluate $f'(27)$.2012-04-03
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    sorry @DavidMitra i still don't get it! Can you plz explain more?2012-04-03
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    Explain more *what*? You should be taking the derivative at some point (first, as it turns out). You never took the derivative. So that's a problem.2012-04-03

1 Answers 1

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First find $f'(x)$. This is important, find the general form of the derivative first:

Using the power rule $$ {d\over dx} x^a=ax^{a-1}, $$ we have,

$$f'(x)= {d\over dx}x^{1/3}={\textstyle{1\over3}}x^{{1\over3}-1}={\textstyle{1\over3}} x^{-2/3}={{1\over 3 x^{2/3}}}.$$

So, whatever $x$ is (as long as it isn't zero), $$ f'(x)={{1\over 3 x^{2/3}}}. $$ Now you know the rule for the function $f'(x)$, and you can evaluate $f'(27)$.

Can you take it from here? (Note $27^{2/3}=(27^{1/3})^2$. )

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    the problem is there is no x in 272012-04-03
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    @SbSangpi The derivative of $f$ is a function. You could be asked for $f'(1)$ or $f'(9)$, or $f'(27)$. That is why you find a formula for $f'(x)$. Once you have that, you can use it to find $f'(x)$ for $x=27$ or for $x$ equal to whatever.2012-04-03
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    $=27^{\frac{2}{3}}$2012-04-03
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    Good answer. The first comment definitely [made me wonder...](http://memegenerator.net/instance/17713910)2012-04-03
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    but my answer is 1/27 not $=27^{\frac{2}{3}}$2012-04-03
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    @SbSangpi $f'(x)={1\over 3x^{2/3}}$; so $f'(27)={1\over 3\cdot 27^{2/3}}={1\over 3\cdot 9}={1\over27}$. This is just evaluating a function ($f'$ here) at a point ($x=27$).2012-04-03
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    hehe..thx! understood now!2012-04-03