Animated Solution for Mathematics - Differentiation: If $x^m\cdot y^n=(x+y{)}^{m+n}$, then $\frac{dy}{dx}$ is

Imagine you are standing before a formidable equation: $x^m\cdot y^n=(x+y{)}^{m+n}$. At first glance, it looks like a tangled mess of variables and exponents.

In the world of JEE Advanced, complexity is often just a mask for elegance. Our goal is to find the derivative $\frac{dy}{dx}$, and the key to unlocking this puzzle lies in the power of logarithms.

When you see variables trapped in exponents, your first instinct should be to liberate them. We apply the natural logarithm, $\ln$, to both sides of the equation:

By using the fundamental properties of logarithms, specifically $\ln (ab)=\ln a+\ln b$ and $\ln (a^b)=b\ln a$, we can transform this product into a simple sum:

Suddenly, the exponents are gone, and we are left with a much more manageable algebraic expression. This is the first step in our journey: simplifying the landscape before we start the climb.

Now that we have simplified our equation, we move to the core of the problem: differentiation. We differentiate both sides with respect to $x$.

Remember, $y$ is not just a constant; it is a function of $x$. This means that whenever we differentiate a term involving $y$, we must invoke the chain rule.

The derivative of $m\ln x$ is simply $\frac{m}{x}$. For the term $n\ln y$, the chain rule gives us $\frac{n}{y}\cdot \frac{dy}{dx}$. On the right side, the derivative of $(m+n)\ln (x+y)$ becomes:

Our equation now stands as:

We want to isolate $\frac{dy}{dx}$, so we expand the right side and move all terms containing $\frac{dy}{dx}$ to the left, and everything else to the right. This is the algebraic dance—a series of careful steps where precision is your best friend.

After rearranging, we have:

By finding a common denominator and simplifying the fractions, we witness a beautiful cancellation. The terms $nx-my$ appear on both sides and vanish, leaving us with the elegant result:

This result is profound. It tells us that for any values of $m$ and $n$, the slope of the tangent at any point $(x,y)$ is simply the ratio of the coordinates.

Geometrically, this means the relation represents straight lines passing through the origin, $y=kx$. The complexity of the original exponents was merely a distraction from this simple, linear truth.

This is the beauty of mathematics—finding the simple, elegant truth hidden beneath the surface of a complex problem. Keep practicing, keep visualizing, and you will find that even the most daunting problems have a simple, beautiful core.