/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 208 $$ \text { Differentiate } \si... [FREE SOLUTION] | 91Ó°ÊÓ

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

Short Answer

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

Step by step solution

01

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 \( \frac{1}{\sqrt{1 - (\frac{1-x}{1+x})^2}} \).
02

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 \( \frac{(1+x) - (1-x)}{(1+x)^2} = \frac{2x}{(1+x)^2} \).
03

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 \( 2 \cdot \sqrt{x} \) .
04

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 \( \sqrt{x} \).

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