What is the derivative of this function $y = {sin}^{- 1} (\frac{3}{t^{2}})$ $y=sin^-1(3/t^2)$ ?

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Answer 1 · 2017-02-20T21:46:59

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$y = arcsin (\frac{1}{3} t^{2})$ $y=arcsin (1/3t^2)$

$sin y = \frac{1}{3} t^{2}$ $siny=1/3t^2$

$cos y \frac{d y}{d x} = \frac{2}{3} t$ $cosy dy/dx=2/3t$

$\frac{d y}{d x} = \frac{2 t}{3 cos y} = \frac{2 t}{3 \sqrt{1 - \frac{1}{9} t^{4}}}$ $dy/dx=(2t)/(3cosy)=(2t)/(3sqrt(1-1/9t^4))$

What is the derivative of this function y=sin^-1(3/t^2)y=sin−1(3t2)y=sin^-1(3/t^2)?