Question

If $x^p{y}^q=(x+y{)}^{p+q}$ ,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
$x^p\frac{d}{dx}{y}^p+y^q\frac{d}{dx}\left(x^p\right)$
$=\frac{d}{dx}{\left(x+y\right)}^{p+q}$
$\Rightarrow x^{p}q{y}^{q-1}\frac{dy}{dx}+y^{q}p{x}^{p-1}$
$=\left(p+q\right){\left(x+y\right)}^{p+q-1}\times \left(1+\frac{dy}{dx}\right)$
$\Rightarrow \frac{q{x}^p{y}^q}{y}\frac{dy}{dx}+\frac{p{y}^q{x}^p}{x}$
$=\left(p+q\right){\left(x+y\right)}^{p+q-1}\times \left(1+\frac{dy}{dx}\right)$
$\Rightarrow x^p{y}^q\left(\frac{q}{y}\frac{dy}{dx}+\frac{p}{x}\right)$
$=\frac{\left(p+q\right){\left(x+y\right)}^{p+q}}{x+y}\times \left(1+\frac{dy}{dx}\right)$
$\Rightarrow x^p{y}^q\left(\frac{q}{y}\frac{dy}{dx}+\frac{p}{x}\right)$
$=\frac{\left(p+q\right)x^p{y}^q}{x+y}\left(1+\frac{dy}{dx}\right)$
$\Rightarrow \frac{q}{y}\frac{dy}{dx}+\frac{p}{x}=\frac{p+q}{x+y}\left(1+\frac{dy}{dx}\right)$
$\Rightarrow \frac{dy}{dx}\left(\frac{q}{y}-\frac{p+q}{x+y}\right)$
$=\frac{p+q}{x+y}-\frac{p}{x}$
$\Rightarrow \frac{dy}{dx}\left(\frac{qx+qy-py-qy}{y\left(x+y\right)}\right)$
$=\frac{px+qx-px-py}{x\left(x+y\right)}$
$\Rightarrow \frac{dy}{dx}\left(\frac{qx-py}{y}\right)$
$=\frac{qx-py}{x}$
$\Rightarrow \frac{dy}{dx}=\frac{y}{x}$