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We have the function:

$$f(x) = \frac{x^2\sqrt[4]{x^3}}{x^3+2}.$$

I rewrote it as $$f(x) = \frac{x^2{x^{3/4}}}{x^3+2}.$$

After a while of differentiating I get the final answer:

$$f(x)= \frac{- {\sqrt[4]{\left(\frac{1}{4}\right)^{19}} + \sqrt[4]{5.5^7}}}{(x^3+2)^2}$$(The minus isn't behind the four)

But my answer sheet gives a different answer, but they also show a wrong calculation, so I don't know what is the right answer, can you help me with this?

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Try logarithmic functions. That might be easier. – smanoos Oct 30 '12 at 19:01
Try writing $$f(x)=\frac{x^{11/4}}{x^3+2}$$ – Manzano Oct 30 '12 at 19:01
Can I conclude from your suggestions that I have the wrong answer? Also, @smanoos , I have not yet reached that level. – JohnPhteven Oct 30 '12 at 19:03
I cannot see where you get your answer from - is there no function of $x$ in the numerator - just a constant? Your post needs some editing to tidy it up (you need the derivative of $f$ in the last equation, for example), but it really isn't clear what you intended to put (eg what is raised to the power 19?). – Mark Bennet Oct 30 '12 at 19:06
+1 for the question. ;-) – Babak S. Nov 2 '12 at 19:09
up vote 1 down vote accepted

Let $y=\frac{x^2\cdot x^{3/4}}{x^3+2}$ so $y=\frac{x^{11/4}}{x^3+2}$ and therefore $y=x^{11/4}\times(x^3+2)^{-1}$. Now use the product rule of two functions: $$(f\cdot g)'=f'\cdot g+f\cdot g'$$ Here $f(x)=x^{11/4}$ and $g(x)=(x^3+2)^{-1}$. So $f'(x)=\frac{11}{4}x^{7/4}$ and $g'(x)=(-1)(3x^2)(x^3+2)^{-2}$. But thinking of your answer in the body, I cannot see where did it come from.

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Although collecting powers is a good idea as suggested, things are made clearer after taking formal logarithm before differentiating:

$$\ln f=2\ln x+\frac{3}{4}\ln x -\ln(x^3+2)$$ $$\frac{f'}{f}=\frac{2}{x}+\frac{3}{4x}-\frac{3x^2}{x^3+2}=\frac{11}{4x}-\frac{3x^2}{x^3+2}=-\frac{x^3-22}{4x(x^3+3)}$$ $$f'=-\frac{x^\frac{11}{4}(x^3-22)}{4x(x^3+2)^2}$$

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But is my answer correct too? – JohnPhteven Oct 30 '12 at 22:19
it is impossible to tell because apparently you tried to evaluate the derivative at some point but only went half way through – Valentin Oct 31 '12 at 7:25

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