Question Statement

Find $\frac{dy}{dx}$ given that $y=x{\sin }^{-1}x+x{\cos }^{-1}x$.

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Background and Explanation

This problem involves differentiating a function that is the sum of two product terms, each containing an inverse trigonometric function. You will need to apply the product rule along with standard derivatives of inverse sine and cosine functions. A key simplification comes from the identity ${\sin }^{-1}x+{\cos }^{-1}x=\frac{\pi }{2}$.

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Solution

Given the function: $y=x{\sin }^{-1}x+x{\cos }^{-1}x$

Differentiate both sides with respect to $x$ using the product rule $\frac{d}{dx}[u(x)v(x)]=u^{\prime }(x)v(x)+u(x)v^{\prime }(x)$ on each term:

$\begin{array}{rl}\frac{dy}{dx}& =\frac{d}{dx}(x)\cdot {\sin }^{-1}x+x\cdot \frac{d}{dx}({\sin }^{-1}x)+\frac{d}{dx}(x)\cdot {\cos }^{-1}x+x\cdot \frac{d}{dx}({\cos }^{-1}x)\\ & =1\cdot {\sin }^{-1}x+x\cdot \frac{1}{\sqrt{1-x^2}}+1\cdot {\cos }^{-1}x+x\cdot \frac{-1}{\sqrt{1-x^2}}\\ & ={\sin }^{-1}x+\frac{x}{\sqrt{1-x^2}}+{\cos }^{-1}x-\frac{x}{\sqrt{1-x^2}}\end{array}$

Notice that the terms $\frac{x}{\sqrt{1-x^2}}$ and $-\frac{x}{\sqrt{1-x^2}}$ cancel each other out:

$\frac{dy}{dx}={\sin }^{-1}x+{\cos }^{-1}x$

Using the standard inverse trigonometric identity:

$\frac{dy}{dx}=\frac{\pi }{2}$

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Key Formulas or Methods Used

• Product Rule: $\frac{d}{dx}[u(x)v(x)]=u^{\prime }(x)v(x)+u(x)v^{\prime }(x)$
• Derivative of inverse sine: $\frac{d}{dx}({\sin }^{-1}x)=\frac{1}{\sqrt{1-x^2}}$
• Derivative of inverse cosine: $\frac{d}{dx}({\cos }^{-1}x)=\frac{-1}{\sqrt{1-x^2}}$
• Inverse trigonometric identity: ${\sin }^{-1}x+{\cos }^{-1}x=\frac{\pi }{2}$

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Summary of Steps

1. Identify the structure: Recognize $y$ as the sum of two products: $x{\sin }^{-1}x$ and $x{\cos }^{-1}x$
2. Apply product rule: Differentiate each term separately using $\frac{d}{dx}[uv]=u^{\prime }v+u{v}^{\prime }$
3. Substitute derivatives: Replace $\frac{d}{dx}({\sin }^{-1}x)$ with $\frac{1}{\sqrt{1-x^2}}$ and $\frac{d}{dx}({\cos }^{-1}x)$ with $\frac{-1}{\sqrt{1-x^2}}$
4. Simplify: Cancel the opposing terms $\frac{x}{\sqrt{1-x^2}}$ and $-\frac{x}{\sqrt{1-x^2}}$
5. Apply identity: Use ${\sin }^{-1}x+{\cos }^{-1}x=\frac{\pi }{2}$ to obtain the final constant derivative