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 $\frac{y}{x}$

Given, $x^p\cdot y^q={(x+y)}^{p+q}$

taking $\log$ on both sides we get,

$\log (x^p\cdot y^q)=\log \left[{(x+y)}^{p+q}\right]\Rightarrow \log x^p+\log y^q=(p+q)\log (x+y)\Rightarrow p\log x+q\log y=(p+q)\log (x+y)$

Differentiating both sides with respect to $x$ we get,

$p\times \frac{1}{x}+q\times \frac{1}{y}\frac{dy}{dx}=(p+q)\frac{1}{(x+y)}(1+\frac{dy}{dx})\Rightarrow \frac{p}{x}+\frac{q}{y}\frac{dy}{dx}=\frac{(p+q)}{(x+y)}+\left(\frac{p+q}{x+y}\right)\frac{dy}{dx}\Rightarrow \frac{dy}{dx}[\frac{p+q}{x+y}-\frac{q}{y}]=\frac{p}{x}-\left(\frac{p+q}{x+y}\right)\Rightarrow \frac{dy}{dx}\left[\frac{py+qy-qx-qy}{y(x+y)}\right]=\frac{px+py-px-qx}{x(x+y)}\Rightarrow \frac{dy}{dx}\frac{(py-qx)}{y(x+y)}=\frac{py-qx}{x(x+y)}\Rightarrow \frac{dy}{dx}=\frac{y}{x}\therefore \frac{dy}{dx}=\frac{y}{x}.$