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Given the polinomyal $f(x)=\frac{x^3}{(4-x^2)^3}$ find $f'(x)$

So, If I try to derive this, first I must to apply the chain rule in the denominator and then derive of the division (...)

$$f'(x)=\frac{x^3}{3(4-x^2)^2(-2)} = \frac{x^3}{-6(4-x^2)^2}$$

(...)?

Or there is another way to do this?

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  • $\begingroup$ Does your problem requires you to use only the chain rule? $\endgroup$ – Ataraxia May 19 '13 at 23:10
  • $\begingroup$ the exercise says find f'(x) but I don't know how to solve this $\endgroup$ – Tomi May 19 '13 at 23:19
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You're doing wrong. A good way to do this when still not expert in derivatives, is to write $$ f(x)=\frac{g(x)}{h(x)} $$ where $$ g(x)=x^3,\qquad h(x)=(4-x^2)^3. $$

Then, first of all, you have to differentiate the quotient: $$ f'(x)=\frac{g'(x)h(x)-g(x)h'(x)}{(h(x))^2} $$

So you need to compute $g'(x)$ and $h'(x)$. The first one is easy: $$ g'(x)=3x^2. $$ The second one needs the chain rule: $$ h'(x)=3(4-x^2)^2(-2x)=-6x(4-x^2)^2. $$ Now plug in the formula above: \begin{align} f'(x)&= \frac{3x^2\cdot(4-x^2)^3 - x^3\cdot(-6x(4-x^2)^2)}{(4-x^2)^6}\\[2ex] &=\frac{3x^2(4-x^2)^3+6x^4(4-x^2)^2}{(4-x^2)^6}\\[2ex] &=\frac{3x^2(4-x^2)+6x^4}{(4-x^2)^4}\\[2ex] &=\frac{12x^2+3x^4}{(4-x^2)^4}=\frac{3x^2(4+x^2)}{(4-x^2)^4} \end{align}

A different strategy may be noting that $$ f(x)=\left(\frac{x}{4-x^2}\right)^3 $$ so, letting $$g(x)=\frac{x}{4-x^2}$$ you have, by the chain rule, $$ f'(x)=3g(x)^2g'(x) $$ The derivative of $g(x)$ can be computed again with the quotient rule: $$ g'(x)=\frac{1\cdot(4-x^2)-x\cdot(-2x)}{(4-x^2)^2}= \frac{4-x^2+2x^2}{(4-x^2)^2}=\frac{4+x^2}{(4-x^2)^2} $$ and so $$ f'(x)=3\left(\frac{x}{(4-x^2)}\right)^2\frac{4+x^2}{(4-x^2)^2} $$ which gives the same result as before.

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You can probably use quotient rule,

$$h(x)=\dfrac{f(x)}{g(x)}$$

$$h'(x)=\dfrac{f'(x)g(x)-g'(x)f(x)}{(g(x))^2}$$

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You want to use the quotient rule then the chain rule.

$$\frac{d}{dx}\frac{f(x)}{g(h(x))}=\frac{g(h(x))f'(x)-f(x)g'(h(x))h'(x)}{g(h(x))^2} $$

so:

$$\frac{d}{dx}\frac{x^3}{(4-x^2)^3} = \frac{(4-x^2)^3(3x^2)-x^3(3(4-x^2)^3)(-2x)}{(4-x^2)^6} $$

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