Question

If $x^p{y}^q={(x+y)}^{p+q}$, then $\frac{dy}{dx}=$

• A. $\frac{y}{x}$
• B. $\frac{-y}{x}$
• C. $\frac{x}{y}$
• D. $-\frac{x}{y}$

Answer 1

The correct option is A

$\frac{y}{x}$

Explanation for the correct option:

$x^p{y}^q={(x+y)}^{p+q}$

By taking log on the both sides.

$p\log x+q\log y=\left(p+q\right)\log \left(x+y\right)$

Differentiate w.r.t $x$

$\begin{array}{rcl}p\left(\frac{1}{x}\right)+q\left(\frac{1}{y}\right)\left(\frac{dy}{dx}\right)& =& \left(p+q\right)\left(\frac{1}{\left(x+y\right)}\right)\times \left[1+\frac{dy}{dx}\right]\\ \Rightarrow \frac{p}{x}+\frac{q}{y}\left(\frac{dy}{dx}\right)& =& \frac{p+q}{x+y}+\frac{p+q}{x+y}\left(\frac{dy}{dx}\right)\\ \Rightarrow \frac{p}{x}-\frac{p+q}{x+y}& =& \left[\frac{p+q}{x+y}-\frac{q}{y}\right]\frac{dy}{dx}\\ & \Rightarrow & \frac{px+py-px-qx}{x\left(x+y\right)}=\frac{py+qy-qx-qy}{y\left(x+y\right)}\frac{dy}{dx}\\ \Rightarrow \frac{py-qx}{x\left(x+y\right)}& =& \frac{py-qx}{y\left(x+y\right)}\frac{dy}{dx}\\ \Rightarrow \frac{dy}{dx}& =& \frac{y}{x}\end{array}$

Hence, option (A) is correct.