Question

The derivative of \(\sin\left(\tan^{-1} e^{2x}\right)\) with respect to \(x\) is:

• A. \(\frac{2e^{2x}\sin\left(\tan^{-1} e^{2x}\right)}{1 + e^{4x}}\)
• B. \(\frac{2e^{2x}\cos\left(\tan^{-1} e^{2x}\right)}{1 + e^{4x}}\)
• C. \(\frac{2e^{2x}\sin\left(\tan^{-1} e^{2x}\right)}{1 + e^{2x}}\)
• D. \(\frac{2e^{2x}\cos\left(\tan^{-1} e^{2x}\right)}{1 + e^{2x}}\)

Hint:

Use the chain rule and the derivative formula for \(\tan^{-1} x\), which is \(\frac{1}{1+x^2}\).

Answer 1

The Correct Option is B

Let \[ y = \sin\left(\tan^{-1} e^{2x}\right) \] Step 1: Use the chain rule for differentiation: \[ \frac{dy}{dx} = \cos\left(\tan^{-1} e^{2x}\right) \cdot \frac{d}{dx} \left(\tan^{-1} e^{2x}\right) \] Step 2: Differentiate \(\tan^{-1} e^{2x}\) using the chain rule: \[ \frac{d}{dx} \tan^{-1} u = \frac{1}{1+u^2} \frac{du}{dx} \] where \(u = e^{2x}\), so \[ \frac{du}{dx} = 2e^{2x} \] Therefore, \[ \frac{d}{dx} \left(\tan^{-1} e^{2x}\right) = \frac{2e^{2x}}{1 + \left(e^{2x}\right)^2} = \frac{2e^{2x}}{1 + e^{4x}} \] Step 3: Combine the results: \[ \frac{dy}{dx} = \cos\left(\tan^{-1} e^{2x}\right) \cdot \frac{2e^{2x}}{1 + e^{4x}} = \frac{2e^{2x} \cos\left(\tan^{-1} e^{2x}\right)}{1 + e^{4x}} \]