Animated Solution for Mathematics - Differentiation: If $f(x)=\{\begin{array}{ll}{x}^3\sin (\frac{1}{x}),& x\ne 0\\ 0,& x=0\end{array}$ then

Welcome, future engineer. Today, we are going to dissect a function that is a classic in the halls of JEE Advanced mathematics: $f(x)=x^3\sin (\frac{1}{x})$.

At first glance, it looks like a simple product, but it hides a beautiful, oscillating complexity. Imagine a curve that is being squeezed tighter and tighter as it approaches the origin, trapped between the envelopes of $y=x^3$ and $y=-x^3$.

Our mission is to find the second derivative of this function at a specific point, $x=\frac{2}{\pi }$. This is not just a calculation; it is a test of your algebraic stamina and your ability to maintain clarity amidst a storm of chain rules.

The First Derivative - The Product Rule Partnership

To find the second derivative, we must first conquer the first. We have a product of two functions: $u=x^3$ and $v=\sin (\frac{1}{x})$. The Product Rule, $\frac{d}{dx}[u\cdot v]=u^{\prime }v+u{v}^{\prime }$, is our best friend here.

Let us differentiate carefully. The derivative of $u=x^3$ is straightforward: $u^{\prime }=3x^2$.

Now, for $v=\sin (\frac{1}{x})$, we invoke the Chain Rule. The derivative of $\sin (\theta )$ is $\cos (\theta )$, and the derivative of the inner function $\frac{1}{x}$ is $-\frac{1}{x^2}$. Thus, $v^{\prime }=\cos (\frac{1}{x})\cdot (-\frac{1}{x^2})$.

Putting these into our formula, we get:

Look at that second term. The $x^3$ and the $\frac{1}{x^2}$ interact beautifully. They simplify, leaving us with a much cleaner first derivative:

The Second Derivative - The Test of Stamina

Now, we must differentiate again. This is where most students stumble, not because the calculus is hard, but because the bookkeeping becomes heavy. We have two terms, and both are products.

For the first term, $\frac{d}{dx}[3x^2\sin (\frac{1}{x})]$: Using the product rule, we get $(6x)\sin (\frac{1}{x})+(3x^2)[\cos (\frac{1}{x})\cdot (-\frac{1}{x^2})]$. Simplifying this, the $x^2$ terms cancel, leaving us with $6x\sin (\frac{1}{x})-3\cos (\frac{1}{x})$.

For the second term, $\frac{d}{dx}[x\cos (\frac{1}{x})]$: Again, the product rule: $(1)\cos (\frac{1}{x})+(x)[-\sin (\frac{1}{x})\cdot (-\frac{1}{x^2})]$. Notice the double negative here! The two negative signs cancel out, leaving us with $\cos (\frac{1}{x})+\frac{1}{x}\sin (\frac{1}{x})$.

Now, we combine them, remembering to subtract the entire second derivative expression:

Distributing the negative sign and grouping like terms, we arrive at the final expression for the second derivative:

The Evaluation - The Moment of Clarity

We have survived the algebra. Now, we evaluate at $x=\frac{2}{\pi }$. Notice the brilliance of this choice of point. When we plug in $x=\frac{2}{\pi }$, the term $\frac{1}{x}$ becomes $\frac{\pi }{2}$.

We know that $\sin (\frac{\pi }{2})=1$ and $\cos (\frac{\pi }{2})=0$. This is the moment where the complexity collapses into simplicity. The cosine term vanishes entirely!

Substituting the values:

Finding a common denominator of $2\pi$, we get the final result:

We have navigated the product rule, the chain rule, and the sign-management traps. The key takeaway here is not just the final answer, but the process: use brackets, simplify early, and trust the math.

When you face these problems in the exam, remember this: the complexity is often just a mask for a beautiful, simple result waiting to be revealed. The final answer is $\frac{24-{\pi }^2}{2\pi }$.