Question

The derivative \( \frac{d}{dx} [\cos(\log x + e^x)] \) at \( x = 1 \) is:

• A. \( -\sin e \)
• B. \( \sin e \)
• C. \( -(1+e)\sin e \)
• D. \( (1+e)\sin e \)

Hint:

To differentiate composite functions, use the chain rule step by step: 1. Differentiate the outer function while keeping the inner function intact. 2. Multiply by the derivative of the inner function.

Answer 1

The Correct Option is C

Step 1: Differentiate the given function. The given function is: \[ y = \cos(\log x + e^x). \] Using the chain rule, the derivative is: \[ \frac{dy}{dx} = -\sin(\log x + e^x) \cdot \frac{d}{dx}(\log x + e^x). \] Step 2: Differentiate \( \log x + e^x \). The derivative of \( \log x \) is \( \frac{1}{x} \), and the derivative of \( e^x \) is \( e^x \). Therefore: \[ \frac{d}{dx}(\log x + e^x) = \frac{1}{x} + e^x. \] Step 3: Substitute into the derivative. Substituting \( \frac{d}{dx}(\log x + e^x) \) into the expression for \( \frac{dy}{dx} \), we get: \[ \frac{dy}{dx} = -\sin(\log x + e^x) \cdot \left( \frac{1}{x} + e^x \right). \] Step 4: Evaluate the derivative at \( x = 1 \). At \( x = 1 \): \[ \log x = \log 1 = 0, \quad e^x = e^1 = e, \quad \frac{1}{x} = \frac{1}{1} = 1. \] Substituting these values: \[ \frac{dy}{dx} = -\sin(0 + e) \cdot (1 + e) = -(1+e)\sin e. \] Thus, the derivative at \( x = 1 \) is \( -(1+e)\sin e \).