Question

If $y=2^{x}\cdot3^{2x-1}$, then $\frac{dy}{dx}$ is equal to

• A. $(log \,2)(log\, 3)$
• B. $(log \,18)$
• C. $(log \,18^2)y^2$
• D. $y(log \,18)$

Answer 1

Answer: (D)

$y(log \,18)$

Answer 2

Given, $y = 2^x \cdot 3^{2x-1}$ Differentiating $w$.$r$.$t$. $x$, we get $\frac{dy}{dx}=2^{x}\cdot\frac{d}{dx}\left(3^{2x-1}\right)+\left(3^{2x-1}\right)+\left(3^{2x-1}\right) \frac{d}{dx}\left(2^{x}\right)\,\ldots\left(i\right)$ Let $3^{2x-1}=u$ $\Rightarrow logu=\left(2x-1\right)log3$ $\Rightarrow \frac{du}{dx}=3^{2x-1}\times2\cdot log3$ $\therefore$ From $(i)$, we have $\frac{dy}{dx}=2^{x}\cdot3^{2x-1}\left(2\right)log\,3+2^{x}\cdot3^{2x-1}\,log\,2$ $\Rightarrow \frac{dy}{dx}=2^{x}\cdot3^{2x-1}\left[2\,log\,3+log\,2\right]$ $\Rightarrow \frac{dy}{dx}=y\,log\,18$