NIMCET 2022 — Mathematics PYQ

To differentiate equations of the form $x^m{y}^n=(x+y{)}^{m+n}$, we use Logarithmic Differentiation.

1. Take the Natural Logarithm ($\ln$) on both sides:

$$\ln (x^m{y}^n)=\ln ((x+y{)}^{m+n})$$

Using the property $\ln (ab)=\ln a+\ln b$ and $\ln (a^b)=b\ln a$:

$$m\ln x+n\ln y=(m+n)\ln (x+y)$$

2. Differentiate both sides with respect to $x$:

$$\frac{d}{dx}(m\ln x+n\ln y)=\frac{d}{dx}((m+n)\ln (x+y))$$

$$\frac{m}{x}+\frac{n}{y}\frac{dy}{dx}=(m+n)\cdot \frac{1}{x+y}\cdot \left(1+\frac{dy}{dx}\right)$$

3. Rearrange the terms to solve for $\frac{dy}{dx}$:

$$\frac{m}{x}+\frac{n}{y}\frac{dy}{dx}=\frac{m+n}{x+y}+\frac{m+n}{x+y}\frac{dy}{dx}$$

Group the $\frac{dy}{dx}$ terms:

$$\left(\frac{n}{y}-\frac{m+n}{x+y}\right)\frac{dy}{dx}=\frac{m+n}{x+y}-\frac{m}{x}$$

Simplify both sides:

$$\left(\frac{n(x+y)-y(m+n)}{y(x+y)}\right)\frac{dy}{dx}=\frac{x(m+n)-m(x+y)}{x(x+y)}$$

$$\left(\frac{nx+ny-my-ny}{y(x+y)}\right)\frac{dy}{dx}=\frac{mx+nx-mx-my}{x(x+y)}$$

$$\left(\frac{nx-my}{y}\right)\frac{dy}{dx}=\frac{nx-my}{x}$$

Canceling $(nx-my)$ from both sides, we get:

$$\frac{1}{y}\frac{dy}{dx}=\frac{1}{x}⟹\frac{dy}{dx}=\frac{y}{x}$$

Correct Option: D) $\frac{y}{x}$