Question

If \( y = \sin^{-1} \left( \frac{2x}{1 + x^2} \right) \), then find \( \frac{dy}{dx} \).

Hint:

To differentiate inverse sine functions, use the chain rule and apply the formula \( \frac{d}{dx} \sin^{-1}(u) = \frac{1}{\sqrt{1 - u^2}} \cdot \frac{du}{dx} \).

Answer 1

Step 1: Differentiation of inverse sine.
We have \( y = \sin^{-1} \left( \frac{2x}{1 + x^2} \right) \). To differentiate this, we will use the chain rule and the derivative formula for \( \sin^{-1} u \), which is: \[ \frac{d}{dx} \sin^{-1}(u) = \frac{1}{\sqrt{1 - u^2}} \cdot \frac{du}{dx} \] where \( u = \frac{2x}{1 + x^2} \).

Step 2: Differentiate \( u \).
First, differentiate \( u = \frac{2x}{1 + x^2} \) using the quotient rule: \[ \frac{du}{dx} = \frac{(1 + x^2)(2) - (2x)(2x)}{(1 + x^2)^2} = \frac{2 + 2x^2 - 4x^2}{(1 + x^2)^2} = \frac{2 - 2x^2}{(1 + x^2)^2} \]

Step 3: Substitute in the formula.
Now, use the derivative formula: \[ \frac{dy}{dx} = \frac{1}{\sqrt{1 - \left( \frac{2x}{1 + x^2} \right)^2}} \cdot \frac{2 - 2x^2}{(1 + x^2)^2} \] Simplifying the term inside the square root: \[ 1 - \left( \frac{2x}{1 + x^2} \right)^2 = \frac{(1 + x^2)^2 - 4x^2}{(1 + x^2)^2} = \frac{1 + 2x^2 + x^4 - 4x^2}{(1 + x^2)^2} = \frac{1 - 2x^2 + x^4}{(1 + x^2)^2} \]

Step 4: Conclusion.
Thus, the derivative is: \[ \frac{dy}{dx} = \frac{2 - 2x^2}{(1 + x^2)^2 \sqrt{\frac{1 - 2x^2 + x^4}{(1 + x^2)^2}}} \]