Question

If $y=2^x\cdot 3^{2x-1}$, then $\frac{dy}{dx}$ is equal to

• A. $(log2)(log3)$
• B. $(log18)$
• C. $(log18^2)y^2$
• D. $y(log18)$

Answer 1

Answer: (D)

Given, $y=2^x\cdot 3^{2x-1}$
Differentiating $w$.$r$.$t$. $x$, we get
$\frac{dy}{dx}=2^x\cdot \frac{d}{dx}\left(3^{2x-1}\right)+\left(3^{2x-1}\right)+\left(3^{2x-1}\right)\frac{d}{dx}\left(2^x\right)\dots \left(i\right)$
Let $3^{2x-1}=u$
$\Rightarrow logu=\left(2x-1\right)log3$
$\Rightarrow \frac{du}{dx}=3^{2x-1}\times 2\cdot log3$
$\therefore$ From $(i)$, we have
$\frac{dy}{dx}=2^x\cdot 3^{2x-1}\left(2\right)log3+2^x\cdot 3^{2x-1}log2$
$\Rightarrow \frac{dy}{dx}=2^x\cdot 3^{2x-1}\left[2log3+log2\right]$
$\Rightarrow \frac{dy}{dx}=ylog18$