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If $u=x^3$ and $v=\arcsin(x)$, prove that $$\frac{du}{dv}=3\sqrt{u(u^{1/3}-u)}$$

I have tried $$\frac{du}{dv}=\frac{\frac{du}{dx}}{\frac{dv}{dx}}$$ but I can't prove it.

Lord_Farin
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user80551
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  • "To prove" questions mean that the answer is steps leading to the equation asked to be proven. – user80551 Jun 29 '13 at 19:21
  • Your first step looks good – where do you get stuck? – Ben Millwood Jun 29 '13 at 19:22
  • Please, try to make the title of your question more informative. E.g., Why does $a\le b$ imply $a+c\le b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. – Lord_Farin Jun 29 '13 at 19:34

2 Answers2

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$$u=x^3 \implies \dfrac{du}{dx}=3x^2$$

$$v=\sin^{-1}x \implies \dfrac{dv}{dx}=\dfrac{1}{\sqrt{1-x^2}}$$

\begin{align} \dfrac{du}{dv}&=\dfrac{\frac{du}{dx}}{\frac{dv}{dx}}\\&=\dfrac{(3x^2)}{\left(\frac{1}{\sqrt{1-x^2}}\right)}\\&=3x^2\sqrt{1-x^2}\\&=3\sqrt{x^4-x^6}\\&=3\sqrt{x^3(x-x^3)}\\&=3\sqrt{u(u^{1/3}-u)} \end{align}

Maazul
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HINT:

$$3\sqrt{u(u^{\frac13}-1)}=3\sqrt{x^3(x-x^3)}=3x^2\sqrt{1-x^2}$$

$$\frac{du}{dx}=3x^2\text{ and } \frac{dv}{dx}=\frac1{\sqrt{1-x^2}}$$

lab bhattacharjee
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