Problem 208 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

$$ \text { Differentiate } \sin ^{-1} \frac{1-x}{1+x} \text { w.r.t. } \sqrt{x} $$

Step-by-Step Solution

Answer

\( \frac{4x\sqrt{x}}{(1+x)^2 \sqrt{1 - (\frac{1-x}{1+x})^2}} \)

Step 1: Differentiate the Outer Function

Differentiate the inverse sine function using its derivative: \( \frac{d}{dx} \sin^{-1}(u) = \frac{1}{\sqrt{1 - u^2}} \cdot u' \). So, the derivative of \( \sin ^{-1} \frac{1-x}{1+x} \) with respect to \( \frac{1-x}{1+x} \) is \( \frac{1}{\sqrt{1 - (\frac{1-x}{1+x})^2}} \).

Step 2: Differentiate the Inner Function

Now find the derivative of the inner function \( \frac{1-x}{1+x} \) with respect to \( x \). We would use the quotient rule which states that \( (f/g)' = f'g - fg' / g^2 \). Applying this gives \( \frac{(1+x) - (1-x)}{(1+x)^2} = \frac{2x}{(1+x)^2} \).

Step 3: Differentiate the Outer Function with respect to \( \sqrt{x} \)

Now, we differentiate \( x \) with respect to \( \sqrt{x} \). We can express \( x \) as \( \sqrt{x}^2 \), and by applying chain rule, we get the derivative as \( 2 \cdot \sqrt{x} \) .

Step 4: Piece it all together

Multiply all the derivative parts together: \( \frac{1}{\sqrt{1 - (\frac{1-x}{1+x})^2}} \cdot \frac{2x}{(1+x)^2} \cdot 2\sqrt{x} \). Simplifying the expression, we get: \( \frac{4x\sqrt{x}}{(1+x)^2 \sqrt{1 - (\frac{1-x}{1+x})^2}} \). This is the derivative of the given function with respect to \( \sqrt{x} \).