Question

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

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

Answer 1

The correct option is D $\frac{y}{x}$

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

Taking log on both sides,
$p\log x+q\log y=(p+q)\log (x+y)$

On differentiating w.r.t. x, we get
$\frac{p}{x}+\frac{q}{y}\cdot \frac{dy}{dx}=\frac{(p+q)}{(x+y)}(1+\frac{dy}{dx})$
$\{\frac{p}{x}-\frac{p+q}{x+y}\}=\{\frac{p+q}{x+y}-\frac{q}{y}\}\frac{dy}{dx}$
$\left\{\frac{px+py-px-qx}{x(x+y)}\right\}=\left\{\frac{py+qy-qx-qy}{y(x+y)}\right\}\frac{dy}{dx}$
$\Rightarrow \frac{(py-qx)}{x}=\frac{(py-qx)}{y}\cdot \frac{dy}{dx}$
$\Rightarrow \frac{dy}{dx}=\frac{y}{x}$