Question

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

Explanation:

Step1: Differentiate \(x\sin^{-1}x\) using product rule

The product rule is \((uv)^\prime = u^\prime v+uv^\prime\). Let \(u = x\) and \(v=\sin^{-1}x\). Then \(u^\prime=1\) and \(v^\prime=\frac{1}{\sqrt{1 - x^{2}}}\). So \((x\sin^{-1}x)^\prime=\sin^{-1}x+\frac{x}{\sqrt{1 - x^{2}}}\)

Step2: Differentiate \(\sqrt{1 - x^{2}}\) using chain rule

Let \(u = 1 - x^{2}\), then \(y=\sqrt{u}=u^{\frac{1}{2}}\). By chain rule \(\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}\). \(\frac{dy}{du}=\frac{1}{2}u^{-\frac{1}{2}}\) and \(\frac{du}{dx}=-2x\). So \(\frac{d}{dx}(\sqrt{1 - x^{2}})=\frac{-2x}{2\sqrt{1 - x^{2}}}=\frac{-x}{\sqrt{1 - x^{2}}}\)

Step3: Find \(y^\prime\)

\(y^\prime=(x\sin^{-1}x)^\prime+(\sqrt{1 - x^{2}})^\prime=\sin^{-1}x+\frac{x}{\sqrt{1 - x^{2}}}+\frac{-x}{\sqrt{1 - x^{2}}}\)

Answer:

\(y^\prime=\sin^{-1}x\)