Question

If \[ y=(e^{2x}-4)(6e^{2x}-5e^{x}+1), \] then \[ \left(\frac{dy}{dx}\right)_{x=0} - \left(\frac{d^{2}y}{dx^{2}}\right)_{x=0} = \ ? \]

• A. \(0\)
• B. \(-5\)
• C. \(4\)
• D. \(6\)

Hint:

For exponential expressions, expanding first often reduces the amount of product-rule differentiation required.

Answer 1

The Correct Option is C

Concept: When values of first and second derivatives are required at a specific point, it is often convenient to first simplify the function and then differentiate term by term. The exponential function satisfies \[ \frac{d}{dx}(e^{ax})=ae^{ax}. \]

Step 1: Expand the given expression.
Given \[ y=(e^{2x}-4)(6e^{2x}-5e^x+1). \] Multiplying, \[ y=6e^{4x}-5e^{3x}+e^{2x}-24e^{2x}+20e^x-4. \] Therefore, \[ y=6e^{4x}-5e^{3x}-23e^{2x}+20e^x-4. \]

Step 2: Find the first derivative.
Differentiating term by term, \[ \frac{dy}{dx} = 24e^{4x} -15e^{3x} -46e^{2x} +20e^x. \] Putting \(x=0\), \[ \left(\frac{dy}{dx}\right)_{x=0} = 24-15-46+20. \] \[ = -17. \]

Step 3: Find the second derivative.
Differentiating again, \[ \frac{d^2y}{dx^2} = 96e^{4x} -45e^{3x} -92e^{2x} +20e^x. \] At \(x=0\), \[ \left(\frac{d^2y}{dx^2}\right)_{x=0} = 96-45-92+20. \] \[ =-21. \]

Step 4: Calculate the required quantity.
\[ \left(\frac{dy}{dx}\right)_{x=0} - \left(\frac{d^2y}{dx^2}\right)_{x=0} \] \[ = (-17)-(-21). \] \[ =4. \] Hence \[ \boxed{4}. \]