CUET · MATHS · PYQ PAPER 2023
The derivative of \(\sin \left(\tan ^{-1} e^{2 x}\right)\) with respect to x is:
- A \(\frac{2 e^{2 x} \sin \left(\tan ^{-1} e^{2 x}\right)}{1+e^{4 x}}\)
- B \(\frac{2 e^{2 x} \cos \left(\tan ^{-1} e^{2 x}\right)}{1+e^{4 x}}\)
- C \(\frac{2 e^{2 x} \sin \left(\tan ^{-1} e^{2 x}\right)}{1+e^{x^2}}\)
- D \(\frac{2 e^{2 x} \cos \left(\tan ^{-1} e^{2 x}\right)}{1+e^{2 x}}\)
Answer & Solution
Correct Answer
(B) \(\frac{2 e^{2 x} \cos \left(\tan ^{-1} e^{2 x}\right)}{1+e^{4 x}}\)
Step-by-step Solution
Detailed explanation
\( \frac{d}{dx} \left(\sin \left(\tan ^{-1} e^{2 x}\right)\right) = \cos \left(\tan ^{-1} e^{2 x}\right) \cdot \frac{d}{dx} \left(\tan ^{-1} e^{2 x}\right) \)…
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