Question:
Differential coefficient of see (\(tan^{-1} x\)) with respect to x is:
Differential coefficient of see (\(tan^{-1} x\)) with respect to x is:
Updated On: May 18, 2024
- \(x\sqrt{1 + x^2}\)
- \(\frac{1}{\sqrt{1 + x^2}}\)
- \(\frac{x}{\sqrt{1 + x^2}}\)
- \(\frac{x}{1 + x^2}\)
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The Correct Option is C
Solution and Explanation
The correct option is (C): \(\frac{x}{\sqrt{1 + x^2}}\)
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