Animated Solution for Mathematics - Differentiation: If $(a+bx)e^{y/x}=x$, then prove that $x^3\frac{d^{2}y}{d{x}^2}=(x\frac{dy}{dx}-y{)}^2$.

Unraveling the Proof

Welcome, future engineer. Today, we are not just solving a problem; we are engaging in a beautiful, logical dance. Calculus proofs in JEE Advanced can often feel like a labyrinth—you start with a simple equation, and suddenly you are lost in a forest of derivatives.

But here is the secret: the math is not trying to trick you. It is trying to guide you. Let us walk through this proof together, step by step, and see how the pieces fall into place.

The Strategic Isolation

We begin with the given equation: $(a+bx)e^{y/x}=x$.

Your first instinct might be to jump straight into the product rule. Stop. Take a breath. If we differentiate $(a+bx)e^{y/x}$ as a product, we are going to generate a massive, tangled expression.

Instead, let us be strategic. We want to isolate the exponential term. By moving $(a+bx)$ to the other side, we get:

This is much cleaner. We have transformed a product rule nightmare into a quotient rule opportunity. This is the first mark of a seasoned problem solver: simplifying the landscape before you start the climb.

The First Derivative

Now, we differentiate both sides with respect to $x$. On the left, we use the chain rule. The derivative of $e^{y/x}$ is $e^{y/x}\cdot \frac{d}{dx}(\frac{y}{x})$.

Applying the quotient rule to $\frac{y}{x}$, we get:

On the right side, we apply the quotient rule to $\frac{x}{a+bx}$. The derivative is:

Look at that! The $bx$ terms cancel out perfectly. This is not a coincidence; it is the math rewarding you for your clean setup. Now, we equate the two sides:

The "Aha!" Moment

Here is where many students panic. They see the $e^{y/x}$ and think, "How do I get rid of this?" Remember our original equation? We know that $e^{y/x}=\frac{x}{a+bx}$.

Let us substitute that back in. This is the bridge between the first derivative and the final proof:

Watch the magic happen. We can cancel one $x$ from the numerator and denominator on the left, and one factor of $(a+bx)$ from both sides. We are left with:

Multiply both sides by $x(a+bx)$, and we get a beautiful, simplified relationship:

The Second Derivative and Victory

We are almost there. We need the second derivative, so let us differentiate this new equation with respect to $x$. On the left side, we use the product rule on $x\frac{dy}{dx}$:

The $\frac{dy}{dx}$ terms cancel out! It is elegant, isn't it? Now, for the right side, we differentiate $\frac{ax}{a+bx}$ using the quotient rule again:

So, we have $x\frac{d^{2}y}{d{x}^2}=\frac{a^2}{(a+bx{)}^2}$.

Finally, look at our expression from Phase 3: $x\frac{dy}{dx}-y=\frac{ax}{a+bx}$. If we divide by $x$, we get $\frac{x\frac{dy}{dx}-y}{x}=\frac{a}{a+bx}$. Squaring this gives us exactly $\frac{a^2}{(a+bx{)}^2}$.

Therefore, we can substitute:

Multiply by $x^2$, and we arrive at the destination:

We have arrived. The proof is complete. Remember, calculus is not about memorizing steps; it is about recognizing patterns. You did great today.