Find $y = ({sin}^{- 1} x)^{x^{2}}$ , then $y^{'} (0) = ?$

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Solution

$y = ({sin}^{- 1} x)^{x^{2}}$ Taking log of both sides, we
get $log y = x^{2} log {sin}^{- 1} x .$
Differentiate both
sides w.r.t. $x$ $\frac{1}{y} \frac{d y}{d x} = 2 x log {sin}^{- 1} x + x^{2} . \frac{1}{{sin}^{- 1} x} \frac{1}{(1 - x^{2})}$
$∴ \frac{d y}{d x} = y [2 x log {sin}^{- 1} x + \frac{x^{2}}{(1 - x^{2}) {sin}^{- 1} x}] = ({sin}^{- 1} x)^{x^{2}} [2 x log {sin}^{- 1} x + \frac{x^{2}}{(1 - x^{2}) {sin}^{- 1} x}]$
$∴ y^{'} (0) = 0$

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