differentiation

Related Questions

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• If $x = {\sin ^{ - 1}}(3t - 4{t^3})$ and $y = {\cos ^{ - 1}}\,\,\sqrt {(1 - {t^2})} $, then $\frac{{dy}}{{dx}}$ is equal to

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• $\frac{d}{{dx}}\left( {{{\tan }^{ - 1}}\frac{{\cos x}}{{1 + \sin x}}} \right) = $

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• $\frac{d}{{dx}}\{ {(\sin x)^x}\} $=

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