Question

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

• A. $\frac{y}{x}log\left(\frac{2}{y}\right)$
• B. $\frac{x}{y}log\left(\frac{2}{y}\right)$
• C. $\frac{y}{x}log\left(\frac{y}{2}\right)$
• D. $\frac{x}{y}log\left(\frac{y}{2}\right)$
• E. $\frac{y}{x}log(2y)$

Answer 1

Answer: (A)

Given, $y^x=2^x$
On taking $\log$ on both sides, we get
$x\log y=x\log 2$
On differentiating w.r.t. $x$, we get
$x\times \frac{1}{y}\frac{dy}{dx}+\log y=\log 2$
$\Rightarrow \frac{dy}{dx}=\frac{y}{x}[\log 2-\log y]$
$=\frac{y}{x}\left[\log \frac{2}{y}\right]$