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Question

# Find the derivative of $y=x \sin ^{-1} x+\sqrt{1-x^2}$. Simplify where possible.

Solution

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$y = x\sin^{-1}x+\sqrt{1-x^2}$

Differentiate

$y' = \dfrac{d\left(x\sin^{-1}x \right)}{dx}+\dfrac{d\left( \sqrt{1-x^2}\right)}{dx}$

$y' = \underbrace{\sin^{-1}x\cdot \dfrac{dx}{dx}+x\cdot \dfrac{d\left(\sin^{-1}x \right)}{dx}}_{\text{\color{#4257b2}Using product Rule}}+\underbrace{\dfrac{d\left( \sqrt{1-x^2}\right)}{d\left(1-x^2 \right)}\times \dfrac{d\left(1-x^2 \right)}{dx}}_{\text{\color{#c34632}Using Chain Rule}}$

$y' = \sin^{-1}x\cdot 1+x\cdot \dfrac{1}{\sqrt{1-x^2}}+\dfrac{1}{2\sqrt{1-x^2}}\times \left(-2x \right)$

$y' = \sin^{-1}x+\dfrac{x}{\sqrt{1-x^2}}-\dfrac{x}{\sqrt{1-x^2}}$

$y' = \sin^{-1}x$

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