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The derivative of sin⁡(tan⁡−1e2x)\sin(\tan^{-1} e^{2x}) with respect to xx is:

Solution

✅ Correct Option: 2

Using the chain rule: ddxsin⁡(tan⁡−1e2x)=cos⁡(tan⁡−1e2x)⋅11+(e2x)2⋅2e2x=2e2xcos⁡(tan⁡−1e2x)1+e4x\frac{d}{dx}\sin(\tan^{-1} e^{2x}) = \cos(\tan^{-1} e^{2x}) \cdot \frac{1}{1+(e^{2x})^2} \cdot 2e^{2x} = \frac{2e^{2x} \cos(\tan^{-1} e^{2x})}{1+e^{4x}}.

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