Question:
The derivative of \(tan^{-1}(\frac{\sqrt{1+x^2}-1}{x})\) with respect to \(sin^{-1}(\frac{2x}{1+x^2})\) is...
The derivative of \(tan^{-1}(\frac{\sqrt{1+x^2}-1}{x})\) with respect to \(sin^{-1}(\frac{2x}{1+x^2})\) is...
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Substitute x = tan theta to simplify both functions to multiples of theta.
Updated On: Oct 1, 2026
- \(\frac{1}{2}\)
- \(\frac{1}{4}\)
- \(1\)
- \(2\)
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The Correct Option is B
Solution and Explanation
Step 1: Substitute x = tan theta:
Let \(x=\tan\theta\). Then \(\sqrt{1+x^2}=\sec\theta\).
Step 2: Simplify u:
\(u=\tan^{-1}\dfrac{\sec\theta-1}{\tan\theta}=\tan^{-1}\dfrac{1-\cos\theta}{\sin\theta}=\tan^{-1}\left(\tan\dfrac\theta2\right)=\dfrac\theta2\). So \(u=\dfrac12\tan^{-1}x\).
Step 3: Simplify v:
\(v=\sin^{-1}\dfrac{2x}{1+x^2}=\sin^{-1}(\sin2\theta)=2\theta=2\tan^{-1}x\).
Step 4: Differentiate:
\(\dfrac{du}{dx}=\dfrac12\cdot\dfrac{1}{1+x^2}\) and \(\dfrac{dv}{dx}=\dfrac{2}{1+x^2}\).
\[ \dfrac{du}{dv}=\dfrac{1/2}{2}=\dfrac14 \]
Step 5: Why the other options are wrong.
1/2 or 2 come from forgetting one of the factors of 2, and 1 would mean \(u\) and \(v\) change equally.
Final Answer:
The derivative is 1/4. \[ \boxed{\text{(B) }\dfrac14} \]
Let \(x=\tan\theta\). Then \(\sqrt{1+x^2}=\sec\theta\).
Step 2: Simplify u:
\(u=\tan^{-1}\dfrac{\sec\theta-1}{\tan\theta}=\tan^{-1}\dfrac{1-\cos\theta}{\sin\theta}=\tan^{-1}\left(\tan\dfrac\theta2\right)=\dfrac\theta2\). So \(u=\dfrac12\tan^{-1}x\).
Step 3: Simplify v:
\(v=\sin^{-1}\dfrac{2x}{1+x^2}=\sin^{-1}(\sin2\theta)=2\theta=2\tan^{-1}x\).
Step 4: Differentiate:
\(\dfrac{du}{dx}=\dfrac12\cdot\dfrac{1}{1+x^2}\) and \(\dfrac{dv}{dx}=\dfrac{2}{1+x^2}\).
\[ \dfrac{du}{dv}=\dfrac{1/2}{2}=\dfrac14 \]
Step 5: Why the other options are wrong.
1/2 or 2 come from forgetting one of the factors of 2, and 1 would mean \(u\) and \(v\) change equally.
Final Answer:
The derivative is 1/4. \[ \boxed{\text{(B) }\dfrac14} \]
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