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Trying to derive an inverse trigonometric function…

I want to find the trigonometric inverse of the following question.

$$\frac{d}{dx}\cos^{-1}(4x^3 - 3x) \quad 0 < x < 1/2 $$

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I don't know what is meant by the inverse of a question. – Gerry Myerson May 15 '11 at 11:50

marked as duplicate by Isaac, t.b., Fabian, Jonas Meyer, J. M. May 16 '11 at 3:28

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Use the chain rule. That is $\displaystyle \frac{d}{dx} f(g(x)) = f'(g(x)) \cdot g'(x)$ and note that $$\frac{\rm d}{ \rm dx} \cos^{-1}{x} = -\frac{1}{\sqrt{1-x^{2}}}$$

ADDED. Put $x = \cos{t}$. Then you want to find the value of $\cos^{-1}(4\cos^{3}{t}-3\cos{t})$. Now you know my trigonometrical identities that $\cos{3\theta} = 4\cos^{3}{\theta} - 3\cos{\theta}$. So the above equation becomes, $\cos^{-1}(\cos{3t})=3t=3\cos^{-1}{x}$. Now Differentiate.

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Any other method then chain rule? – zainab May 16 '11 at 13:19
@Zainab: Yes subsitution method. – user9413 May 16 '11 at 13:41
@Zainab: I hope this is what you were thinking off. Don't forget to accept an answer. – user9413 May 16 '11 at 14:31

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