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: Rewrite y
$y=\dfrac{1}{b}(ab^{2})^{x}$, because $a^{x}b^{2x}=(ab^{2})^{x}$.

Step 2: Differentiate twice
For $y=k\,m^{x}$, $y''=k\,m^{x}(\log m)^{2}$.

Step 3: Substitute
With $m=ab^{2}$ this gives $y''=y(\log(ab^{2}))^{2}$.

Final Answer:
Writing y as k m^x gives y'' = y (log m)^2 with m = ab^2. \[ \boxed{\text{(C) }y\left(\log(ab^{2})\right)^{2}} \]