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
$x \times \frac{1}{y} \frac{d y}{d x}+\log y=\log 2$
$\Rightarrow \frac{d y}{d x} =\frac{y}{x}[\log 2-\log y] $
$=\frac{y}{x}\left[\log \frac{2}{y}\right]$