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Question

Find the derivative of y=xsin1x+1x2y=x \sin ^{-1} x+\sqrt{1-x^2}. Simplify where possible.

Solution

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Answered 7 months ago
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y=xsin1x+1x2y = x\sin^{-1}x+\sqrt{1-x^2}

Differentiate

y=d(xsin1x)dx+d(1x2)dxy' = \dfrac{d\left(x\sin^{-1}x \right)}{dx}+\dfrac{d\left( \sqrt{1-x^2}\right)}{dx}

y=sin1xdxdx+xd(sin1x)dxUsing product Rule+d(1x2)d(1x2)×d(1x2)dxUsing Chain Ruley' = \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=sin1x1+x11x2+121x2×(2x)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=sin1x+x1x2x1x2y' = \sin^{-1}x+\dfrac{x}{\sqrt{1-x^2}}-\dfrac{x}{\sqrt{1-x^2}}

y=sin1xy' = \sin^{-1}x

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