Question

If \(y=\mathrm{a}^x \cdot \mathrm{~b}^{2 x-1}\), then \(\frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}\) is equal to

• A. \(y(\mathrm{log}(ab^{2}))\)
• B. \(y^{2}(\mathrm{log}(a\text{ }\mathrm{b}^{2}))\)
• C. \(y(\mathrm{log}(ab^{2}))^{2}\)
• D. \(y^{2}(\mathrm{log}(ab))^{2}\)

Hint:

Take logarithms and differentiate; each derivative multiplies y by the same constant.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
An exponential function has a derivative equal to a constant times itself, so use logarithmic differentiation.

Step 2: First derivative
\(\log y=x\log a+(2x-1)\log b\). Differentiate: \(\dfrac{1}{y}\dfrac{dy}{dx}=\log a+2\log b=\log(ab^{2})\).

Step 3: Second derivative
So \(\dfrac{dy}{dx}=y\log(ab^{2})\). Differentiate again, with \(\log(ab^{2})\) constant: \(\dfrac{d^{2}y}{dx^{2}}=\log(ab^{2})\dfrac{dy}{dx}\).

Step 4: Result
\(\dfrac{d^{2}y}{dx^{2}}=y\,[\log(ab^{2})]^{2}\).

Final Answer:
The second derivative is y times the square of log(ab^2). \[ \boxed{\text{(C) }y\left(\log(ab^{2})\right)^{2}} \]