Q.Find $\frac{dy}{dx}$ in the following: $x^3 + x^2y + xy^2 + y^3 = 81$

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

Differentiate term by term:

• $\frac{d}{dx}(x^3) = 3x^2$
• $\frac{d}{dx}(x^2y) = 2xy + x^2\frac{dy}{dx}$ (product rule)
• $\frac{d}{dx}(xy^2) = y^2 + 2xy\frac{dy}{dx}$ (product rule)
• $\frac{d}{dx}(y^3) = 3y^2\frac{dy}{dx}$
• $\frac{d}{dx}(81) = 0$

So:

$3x^2 + 2xy + x^2\frac{dy}{dx} + y^2 + 2xy\frac{dy}{dx} + 3y^2\frac{dy}{dx} = 0$ …

The equation $x^3 + x^2y + xy^2 + y^3 = 81$ mixes $x$ and $y$ together — you cannot solve for $y$ in terms of $x$ easily (it’s a cubic in $y$). So we use implicit differentiation: assume $y$ is a function of $x$, differentiate both sides with respect to $x$, and treat $y$ as $y(x)$. Every time we hit a $y$, we apply the chain rule: $\frac{d}{dx}(y^n) = n y^{n-1} \frac{dy}{dx}$.

Let’s go term by term.

1. Differentiate $x^3$:

   $\frac{d}{dx}(x^3) = 3x^2$.

2. Differentiate $x^2 y$:

   This is a product of $x^2$ and $y$. Use the product rule:

   $\frac{d}{dx}(x^2 y) = \frac{d}{dx}(x^2) \cdot y + x^2 \cdot \frac{d}{dx}(y) = 2x \cdot y + x^2 \cdot \frac{dy}{dx}$.

3. Differentiate $x y^2$:

   Again a product: $x$ times $y^2$.

   $\frac{d}{dx}(x y^2) = \frac{d}{dx}(x) \cdot y^2 + x \cdot \frac{d}{dx}(y^2) = 1 \cdot y^2 + x \cdot (2y \frac{dy}{dx}) = y^2 + 2xy \frac{dy}{dx}$.

4. Differentiate $y^3$:

   Chain rule: $\frac{d}{dx}(y^3) = 3y^2 \frac{dy}{dx}$.

5. Differentiate the right side:

   $\frac{d}{dx}(81) = 0$.

Now put it all together:

Collect the terms without $\frac{dy}{dx}$ and those with it:

• Terms without $\frac{dy}{dx}$: $3x^2 + 2xy + y^2$.
• Terms with $\frac{dy}{dx}$: $x^2 \frac{dy}{dx} + 2xy \frac{dy}{dx} + 3y^2 \frac{dy}{dx} = (x^2 + 2xy + 3y^2) \frac{dy}{dx}$.

So the equation becomes: …

Method: Implicit Differentiation with Several Mixed-Power Terms

This method handles equations where $x$ and $y$ appear together in more than one mixed term (like $x^2y$ and $xy^2$) — each such term needs its own product-rule expansion before you can collect $\frac{dy}{dx}$.

Steps

Step 1: List every term and classify it

Sort the equation's terms into three kinds: pure-$x$ powers (differentiate normally), pure-$y$ powers (chain rule adds $\frac{dy}{dx}$), and mixed terms like $x^2y$ or $xy^2$ (need the product rule).

Step 2: Expand each mixed term with the product rule

For a term $x^a y^b$, treat it as a product of $x^a$ and $y^b$:

Apply this once per mixed term — several such terms means doing this step several times, one term at a time.

Step 3: Differentiate the remaining pure-power and constant terms

Pure $x$-power terms differentiate normally; a pure $y$-power term $y^n$ differentiates to $ny^{n-1}\frac{dy}{dx}$; a constant differentiates to $0$.

Step 4: Separate "with $\frac{dy}{dx}$" from "without," factor, and solve …

Common Mistakes

Mistake 1: Only partially expanding a mixed term like $x^2y$

It's easy to differentiate $x^2y$ as $2xy$ alone, forgetting the second half of the product rule. Why it's wrong: $x^2y$ is a product of $x^2$ and $y(x)$, so it needs $2xy+x^2\frac{dy}{dx}$ — both the derivative of $x^2$ times $y$, and $x^2$ times the derivative of $y$. Correct approach: apply the product rule fully to each mixed term separately before combining.

Mistake 2: Losing track of a term while collecting four differentiated pieces …