What is the derivative of this function $y = {sin}^{- 1} \sqrt{x}$ $y=sin^-1sqrtx$ ?

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Answer 1 · 2017-03-23T05:48:49

$\frac{d y}{d x} = \frac{1}{2 \sqrt{1 - x} \sqrt{x}}$ $dy/dx= 1/(2sqrt(1-x)sqrt(x)$

$y = {sin}^{- 1} \sqrt{x}$ $y =sin^-1 sqrtx$

Applying the standard differential and the chain rule

$\frac{d y}{d x} = \frac{1}{\sqrt{1 - {(\sqrt{x})}^{2}}} \cdot \frac{1}{2} x^{- \frac{1}{2}}$ $dy/dx= 1/sqrt(1-(sqrtx)^2) * 1/2x^(-1/2)$

$= \frac{1}{\sqrt{1 - x} \cdot 2 \sqrt{x}}$ $= 1/(sqrt(1-x)* 2sqrt(x)$

$= \frac{1}{2 \sqrt{1 - x} \sqrt{x}}$ $= 1/(2sqrt(1-x)sqrt(x)$

What is the derivative of this function y=sin^-1sqrtxy=sin−1√xy=sin^-1sqrtx?