Q.Find $\frac{dy}{dx}$ in the following: $x^2 + xy + y^2 = 100$

Concept: Implicit Differentiation — differentiate both sides with respect to $x$, treating $y$ as a function of $x$, then solve for $\frac{dy}{dx}$.

Step 1: Differentiate term by term:

Step 2: The derivative of the constant $100$ is $0$. So:

$2x + y + x\frac{dy}{dx} + 2y\frac{dy}{dx} = 0.$ …

This is a classic implicit differentiation problem. The equation $x^2 + xy + y^2 = 100$ cannot be easily solved for $y$ in terms of $x$ (you could use the quadratic formula, but it gets messy). Instead, we differentiate both sides as they are, treating $y$ as an unknown function of $x$.

The key idea: whenever you differentiate a term containing $y$, you apply the chain rule. For $y^2$, the derivative is $2y \cdot \frac{dy}{dx}$. For $xy$, you use the product rule: derivative of $x$ times $y$, plus $x$ times the derivative of $y$.

Let’s work through it step by step.

1. Differentiate $x^2$

   The derivative of $x^2$ with respect to $x$ is $2x$.

2. Differentiate $xy$

   This is a product of $x$ and $y$. Using the product rule:

3. Differentiate $y^2$ Here $y$ is a function of $x$, so by the chain rule:

4. Differentiate the right-hand side The constant $100$ differentiates to $0$.

Putting it all together, the derivative of the entire equation is:

Now we solve for $\frac{dy}{dx}$.

5. Collect the $\frac{dy}{dx}$ terms From the expression above, the terms containing $\frac{dy}{dx}$ are $x\frac{dy}{dx}$ and $2y\frac{dy}{dx}$. So:

6. Isolate $\frac{dy}{dx}$ Move the terms without $\frac{dy}{dx}$ to the other side: …

Method: Implicit Differentiation for a Symmetric Equation in $x$ and $y$

Use this method whenever $x$ and $y$ appear mixed in the same equation (here, a symmetric expression like $x^2+xy+y^2$) and solving for $y$ explicitly would be awkward or impossible.

Steps

Step 1: Treat $y$ as $y(x)$ everywhere it appears

Every power of $y$ picks up a $\frac{dy}{dx}$ factor from the chain rule:

Step 2: Use the product rule on the mixed term

A term like $xy$ is a product of $x$ and $y(x)$:

Writing only $x\frac{dy}{dx}$ (dropping the plain $y$) or only $y$ (dropping $x\frac{dy}{dx}$) is incomplete — the product rule always contributes two pieces.

Step 3: Differentiate any constant term to zero

A constant right-hand side (or any constant appearing anywhere) differentiates to $0$ — it never generates a $\frac{dy}{dx}$ term.

Step 4: Group, factor, solve …

Common Mistakes

Mistake 1: Dropping half of the product rule on $xy$

Students often write $\frac{d}{dx}(xy)$ as just $y$ or just $x\frac{dy}{dx}$. Why it's wrong: $xy$ is a product of $x$ and $y(x)$, so the product rule contributes both terms, $y+x\frac{dy}{dx}$. Correct approach: expand every mixed product fully before collecting terms.

Mistake 2: Forgetting the chain-rule factor on $y^2$ …