Question Statement

Find $\frac{dy}{dx}$ given that:

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

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Background and Explanation

This problem involves differentiating a quotient where the numerator is an inverse trigonometric function (${\sin }^{-1}x$) and the denominator is a trigonometric function ($\sin x$). You will need to apply the quotient rule for differentiation, combined with the standard derivatives of inverse sine and sine functions.

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Solution

We are given the function:

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

To differentiate this with respect to $x$, we use the quotient rule. Recall that for a function $y=\frac{u}{v}$, the derivative is:

$\frac{dy}{dx}=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}$

Here, let $u={\sin }^{-1}x$ and $v=\sin x$.

First, identify the required derivatives:

• $\frac{d}{dx}({\sin }^{-1}x)=\frac{1}{\sqrt{1-x^2}}$
• $\frac{d}{dx}(\sin x)=\cos x$

Now apply the quotient rule:

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

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Key Formulas or Methods Used

• Quotient Rule: $\frac{d}{dx}\left(\frac{u}{v}\right)=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}$
• Derivative of inverse sine: $\frac{d}{dx}({\sin }^{-1}x)=\frac{1}{\sqrt{1-x^2}}$
• Derivative of sine: $\frac{d}{dx}(\sin x)=\cos x$

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Summary of Steps

1. Identify the function structure as a quotient $y=\frac{u}{v}$ with $u={\sin }^{-1}x$ and $v=\sin x$
2. State the quotient rule formula for differentiation
3. Differentiate the numerator: $\frac{d}{dx}({\sin }^{-1}x)=\frac{1}{\sqrt{1-x^2}}$
4. Differentiate the denominator: $\frac{d}{dx}(\sin x)=\cos x$
5. Substitute these derivatives into the quotient rule formula
6. Simplify the resulting expression to obtain $\frac{dy}{dx}$