Question:
If \(y = cos^2[cot^{-1}(\sqrt{\frac{1-x}{1+x}})]\) then \(\frac{dy}{dx} =\) ........
If \(y = cos^2[cot^{-1}(\sqrt{\frac{1-x}{1+x}})]\) then \(\frac{dy}{dx} =\) ........
Show Hint
Substitute x = cos 2 theta to simplify the inner expression.
Updated On: Oct 1, 2026
- \(\frac{-1}{2}\)
- \(\frac{1}{2}\)
- \(1\)
- \(-1\)
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Concept
The expression inside has \(\sqrt{\dfrac{1-x}{1+x}}\), which simplifies with the substitution \(x=\cos2\theta\).
Step 2: Substitute
\[ \frac{1-x}{1+x}=\frac{1-\cos2\theta}{1+\cos2\theta}=\tan^2\theta\Rightarrow\sqrt{\frac{1-x}{1+x}}=\tan\theta \]
Step 3: Simplify
\[ \cot^{-1}(\tan\theta)=\frac\pi2-\theta \]
\[ y=\cos^2\left(\frac\pi2-\theta\right)=\sin^2\theta=\frac{1-\cos2\theta}{2}=\frac{1-x}{2} \]
Step 4: Differentiate
\[ \frac{dy}{dx}=-\frac12 \]
This is option (A).
Final Answer:
The expression simplifies to (1 - x)/2, so the derivative is -1/2, option (A). \[ \boxed{-\frac{1}{2}} \]
The expression inside has \(\sqrt{\dfrac{1-x}{1+x}}\), which simplifies with the substitution \(x=\cos2\theta\).
Step 2: Substitute
\[ \frac{1-x}{1+x}=\frac{1-\cos2\theta}{1+\cos2\theta}=\tan^2\theta\Rightarrow\sqrt{\frac{1-x}{1+x}}=\tan\theta \]
Step 3: Simplify
\[ \cot^{-1}(\tan\theta)=\frac\pi2-\theta \]
\[ y=\cos^2\left(\frac\pi2-\theta\right)=\sin^2\theta=\frac{1-\cos2\theta}{2}=\frac{1-x}{2} \]
Step 4: Differentiate
\[ \frac{dy}{dx}=-\frac12 \]
This is option (A).
Final Answer:
The expression simplifies to (1 - x)/2, so the derivative is -1/2, option (A). \[ \boxed{-\frac{1}{2}} \]
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