Animated Solution for Mathematics - Differentiation: Let $f(x)=x\sin \pi x,x>0$. Then for all natural numbers $n,f^{\prime }(x)$ vanishes at

Welcome, fellow traveler on the path of JEE Advanced mathematics. Today, we are not just solving a problem; we are witnessing a beautiful dance between algebra and geometry.

We are given the function $f(x)=x\sin (\pi x)$ and asked to find where its derivative, $f^{\prime }(x)$, vanishes. This is a classic setup that tests not just your ability to differentiate, but your ability to think beyond the page.

The Derivative

First, let us apply our trusty tool, the Product Rule. When we differentiate $f(x)=x\sin (\pi x)$, we treat $x$ as our first function and $\sin (\pi x)$ as our second.

The derivative of $x$ is $1$, and the derivative of $\sin (\pi x)$—thanks to the Chain Rule—is $\pi \cos (\pi x)$. Putting these together, we get:

To find where the derivative vanishes, we set this expression to zero:

This is the moment where many students panic. You look at this equation and think, 'How do I isolate $x$?' The answer is: you don't.

The Graphical Bridge

This is a transcendental equation. It refuses to be solved by simple algebraic manipulation. But in the world of JEE, when algebra hits a wall, geometry opens a door.

Let us rearrange our equation: $\sin (\pi x)=-\pi x\cos (\pi x)$. If we divide both sides by $\cos (\pi x)$, we get the elegant form:

Now, instead of solving for $x$, we are looking for the intersection of two functions: $y_1=\tan (\pi x)$ and $y_2=-\pi x$. The roots of our derivative are simply the $x$-coordinates where these two graphs cross.

The Asymptote Analysis

We are investigating the interval $(n,n+1)$. Here, we must be careful. The tangent function is not continuous everywhere; it has vertical asymptotes at odd multiples of $\pi /2$.

In our interval, this occurs at $x=n+1/2$. This asymptote acts as a wall, splitting our interval into two distinct sub-intervals.

In the first sub-interval, $(n,n+1/2)$, the tangent function $\tan (\pi x)$ is strictly positive, rising from $0$ to $+\infty$. However, our line $y_2=-\pi x$ is strictly negative for all $x>0$. A positive value can never equal a negative value; thus, there is no intersection here.

The Unique Intersection

Now, look at the second sub-interval: $(n+1/2,n+1)$. Here, the tangent function emerges from $-\infty$ and increases strictly until it reaches $0$ at $x=n+1$.

Our line $y_2=-\pi x$ is a steady, decreasing negative value. Because our tangent curve is climbing from the depths of negative infinity up to zero, and our line is sitting there in the negative region, they must cross.

Because one is strictly increasing while the other is strictly decreasing, they can cross exactly once. By the Intermediate Value Theorem, we have found our unique root.

This, my friends, is the elegance of calculus. We didn't need to calculate the exact value of $x$; we only needed to understand the behavior of the functions. You have successfully navigated the trap. Keep this mindset—visualize, analyze, and conquer.