Question

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

• A. $ \frac{y}{x} $
• B. $ \frac{py}{qx} $
• C. $ \frac{x}{y} $
• D. $ \frac{qy}{px} $

Answer 1

Answer: (A)

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