Animated Solution for Mathematics - Differentiation: If $y^2=P(x)$, a polynomial of degree 3, then $2\frac{d}{dx}\left(y^3\frac{d^{2}y}{d{x}^2}\right)$ equals

Welcome, fellow traveler on the path to JEE Advanced mastery. Today, we are not just solving a problem; we are uncovering the hidden symmetry between a simple algebraic relation and the power of calculus.

We are given the equation $y^2=P(x)$, where $P(x)$ is a polynomial of degree 3. Our mission is to evaluate the expression $2\frac{d}{dx}(y^3{y}^{\prime \prime })$.

At first glance, this looks like a daunting task involving high-order derivatives. However, as we peel back the layers, you will see how the structure of the polynomial guides us to a surprisingly elegant solution.

We begin with our anchor: $y^2=P(x)$. In the world of JEE, whenever you see a function defined by a polynomial, you should immediately think about its derivatives.

If $P(x)$ is a cubic, then $P^{\prime }(x)$ is a quadratic, $P^{\prime \prime }(x)$ is linear, and $P^{\prime \prime \prime }(x)$ is a constant. This is our geometric roadmap. We need to find $y^{\prime \prime }$, so let us start by differentiating our anchor with respect to $x$.

Applying the chain rule, the derivative of $y^2$ is $2y{y}^{\prime }$. Thus, we get:

This is our first key relation. It connects the velocity of our function $y$ to the slope of the polynomial $P(x)$.

To reach the second derivative $y^{\prime \prime }$, we must differentiate again. We apply the product rule to the left side of our equation $2y{y}^{\prime }=P^{\prime }(x)$.

The derivative of $2y{y}^{\prime }$ is $2(y^{\prime }{)}^2+2y{y}^{\prime \prime }$. On the right side, the derivative of $P^{\prime }(x)$ is simply $P^{\prime \prime }(x)$. So, we arrive at:

Now, look closely at what we have. We have a term $2y{y}^{\prime \prime }$, but our target expression requires $y^3{y}^{\prime \prime }$.

We need to transform $2y{y}^{\prime \prime }$ into $4y^3{y}^{\prime \prime }$. The most logical step is to multiply the entire equation by $2y^2$. This scales the $2y{y}^{\prime \prime }$ term perfectly to $4y^3{y}^{\prime \prime }$ and allows us to substitute $y^2$ with $P(x)$ later.

Multiplying our equation $2(y^{\prime }{)}^2+2y{y}^{\prime \prime }=P^{\prime \prime }(x)$ by $2y^2$, we get:

Now, let us perform the substitution. We know from our first step that $2y{y}^{\prime }=P^{\prime }(x)$, which implies $4y^2(y^{\prime }{)}^2=(P^{\prime }(x){)}^2$.

Substituting $y^2=P(x)$ and the squared derivative into our equation, we get:

Rearranging to isolate our target term, $4y^3{y}^{\prime \prime }$, we find:

We are now at the threshold of the solution. We need to find $2\frac{d}{dx}(y^3{y}^{\prime \prime })$.

Let us differentiate our isolated expression with respect to $x$. The left side becomes $\frac{d}{dx}(4y^3{y}^{\prime \prime })$, which is $4\frac{d}{dx}(y^3{y}^{\prime \prime })$.

On the right side, we apply the product rule to $2P(x)P^{\prime \prime }(x)$ and the chain rule to $(P^{\prime }(x){)}^2$:

Look at the magic happening here! The terms $2P^{\prime }(x)P^{\prime \prime }(x)$ and $-2P^{\prime }(x)P^{\prime \prime }(x)$ cancel out perfectly. We are left with:

Dividing both sides by 2, we finally obtain the result:

And there it is. The complexity of the derivatives has vanished, leaving us with a clean, elegant relationship.

Because $P(x)$ is a cubic polynomial, $P^{\prime \prime \prime }(x)$ is a constant, making the final expression a simple multiple of the original polynomial. This problem teaches us that in calculus, if you follow the logical steps and trust the algebraic structure, the most intimidating expressions often simplify into something beautiful.