Question

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

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

Answer 1

The correct option is D

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

Explanation for the correct option.

Step 1. Find the value of $\frac{dy}{dx}$.

Differentiate the equation $y=a^x{b}^{2x-1}$ with respect to $x$.

$\begin{array}{rcl}\frac{dy}{dx}& =& \frac{d}{dx}\left(a^x{b}^{2x-1}\right)\\ & =& \left(a^x\log a\right)b^{2x-1}+a^x\left(b^{2x-1}\log b\right)2\\ & =& a^x{b}^{2x-1}\left[\log a+2\log b\right]\end{array}$

Step 2. Find the value of $\frac{d^{2}y}{d{x}^2}$.

Differentiate the equation $\frac{dy}{dx}=a^x{b}^{2x-1}\left[\log a+2\log b\right]$ with respect to $x$.

$\begin{array}{rcl}\frac{d^{2}y}{d{x}^2}& =& \frac{d}{dx}\left[a^x{b}^{2x-1}\left[\log a+2\log b\right]\right]\\ & =& \left[\log a+2\log b\right]\left[\left(a^x\log a\right)b^{2x-1}+a^x\left(b^{2x-1}\log b\right)2\right]\\ & =& a^x{b}^{2x-1}{\left[\log a+2\log b\right]}^2\\ & =& y{\left[\log a+\log b^2\right]}^2\mathbf{}\left[y=a^x{b}^{2x-1},mloga=log{a}^m\right]\\ & =& y{\left(\log a{b}^2\right)}^2\left[loga+logb=logab\right]\end{array}$

Hence, the correct option is D.