Animated Solution for Mathematics - Differentiation: If $f(x)=x{e}^{x(1-x)}$, then $f(x)$ is

Welcome, future engineers! Today, we are going to dissect a function that, at first glance, might look like a daunting beast. We are looking at $f(x)=x{e}^{x(1-x)}$.

When you see a product of a polynomial and an exponential, your instinct might be to panic. But I want you to take a deep breath. In the world of JEE Advanced, complexity is often just a mask for elegance.

Our mission is to find where this function is increasing—that is, where the slope of the tangent is positive.

To understand the behavior of a function, we must look at its rate of change. We need the derivative, $f^{\prime }(x)$. We have a product of two functions: $u=x$ and $v=e^{x-x^2}$.

The Product Rule, our trusty companion, tells us that $\frac{d}{dx}(uv)=u^{\prime }v+u{v}^{\prime }$.

Let's break it down. The derivative of $u=x$ is simply $1$. Now, for $v=e^{x-x^2}$, we must invoke the Chain Rule.

The derivative of $e^g$ is $e^g\cdot g^{\prime }$. Here, $g=x-x^2$, so $g^{\prime }=1-2x$. Thus, the derivative of $v$ is $e^{x-x^2}(1-2x)$.

Putting it all together, we get:

Look at that! The exponential term $e^{x-x^2}$ is common to both parts. This is the moment where the algebra starts to simplify beautifully. We factor it out:

Now, let's clean up the interior of that bracket. We distribute the $x$ to get $1+x-2x^2$. Rearranging this into standard quadratic form, we have $-2x^2+x+1$.

Our derivative now looks like this:

To find the critical points, we need to factor this quadratic. It is often easier to work with a positive leading coefficient, so let's pull out a negative sign:

Splitting the middle term $-x$ into $-2x+x$, we factor the quadratic as $(2x+1)(x-1)$. Our final, elegant derivative is:

This is the climax of our problem. We want to know where $f(x)$ is increasing, which means we need $f^{\prime }(x)\ge 0$.

Recall that $e^{x-x^2}$ is always strictly positive. It never touches the x-axis, and it never dips below it. It is a purely positive scaling factor.

Therefore, it has absolutely no impact on the sign of our derivative. We can effectively ignore it for the purpose of our inequality:

Now, watch closely! We multiply the entire inequality by $-1$. This is the golden rule of inequalities: when you multiply by a negative number, the inequality sign flips!

We have a quadratic expression with a positive leading coefficient, and we want it to be less than or equal to zero. This happens between the roots. Our roots are $x=-\frac{1}{2}$ and $x=1$.

Using the Wavy Curve method, we see that the expression is negative between these two roots. Thus, the function is increasing on the interval $[-\frac{1}{2},1]$.

The Beauty of Calculus

We started with a complex-looking exponential function, and through the systematic application of the Product Rule, Chain Rule, and careful algebraic factoring, we reduced it to a simple interval.

This is the essence of JEE Advanced physics and mathematics—taking a chaotic problem and finding the underlying order. You didn't just solve a problem; you navigated the logic of the curve. Keep this confidence with you. You are ready for the next challenge!