Question

If $y=a^x\cdot b^{2x-1}$, then $\frac{d^{2}y}{d{x}^2}$ is

• A. $y^2\cdot \log a{b}^2$
• B. $y\cdot (\log a{b}^2{)}^2$
• C. $y^2$
• D. $y\cdot (\log a^{2}b{)}^2$

Answer 1

Answer: (B)

$y\cdot (\log a{b}^2{)}^2$

Given, $y=a^x\cdot b^{2x-1}$
Taking log on both sides, we get
$\log y=x\log a+(2x-1)\log b$
$\Rightarrow \frac{1}{y}\frac{dy}{dx}=\log a+2\log b$
$\Rightarrow \frac{dy}{dx}=y\log a{b}^2$
$\therefore \frac{d^{2}y}{d{x}^2}=\frac{dy}{dx}\log a{b}^2=y(\log a{b}^2{)}^2$