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

The key idea is 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:

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

So the equation becomes:

Group all $\frac{dy}{dx}$ terms on one side: …

The equation $xy + y^2 = \tan x + y$ mixes $x$ and $y$ together — you cannot solve for $y$ in terms of $x$ easily (and you don't need to). When $y$ is not written explicitly as $y = f(x)$, we use implicit differentiation: differentiate both sides of the equation with respect to $x$, remembering that $y$ is a function of $x$, so every time you differentiate a $y$-term, you multiply by $\frac{dy}{dx}$ (by the chain rule).

Let’s work through it step by step.

1. Differentiate $xy$ This is a product: $x$ times $y$. Using the product rule:

2. Differentiate $y^2$ Treat $y$ as a function of $x$. By the chain rule:

3. Differentiate $\tan x$ This is a standard derivative:

4. Differentiate $y$ on the right-hand side Again, $y$ is a function of $x$:

Now put it all together, differentiating both sides of the original equation:

5. Collect all $\frac{dy}{dx}$ terms on one side Bring the $\frac{dy}{dx}$ from the right to the left:

6. Factor out $\frac{dy}{dx}$ …

Method: Implicit Differentiation with a Product Term

This method applies whenever $x$ and $y$ are tangled together in an equation — especially when a term like $xy$ multiplies the two variables — so $y$ cannot be written as an explicit formula in $x$ before differentiating.

Steps

Step 1: Mark every $y$-term for the chain rule

Because $y$ secretly means $y(x)$, differentiating any power of $y$ produces an extra factor of $\frac{dy}{dx}$:

Even a bare $y$ (power 1) differentiates to $\frac{dy}{dx}$, never to $0$ — $y$ is not a constant.

Step 2: Apply the product rule wherever $x$ and $y$ multiply

A term such as $xy$ is a product of two functions of $x$, so it needs $\frac{d}{dx}(uv)=u'v+uv'$:

Both pieces are required — dropping either one is the single most common slip in this technique.

Step 3: Differentiate pure-$x$ terms normally

Any term containing only $x$ differentiates by the ordinary rules with no extra factor, e.g. $\frac{d}{dx}(\tan x)=\sec^2 x$.

Step 4: Collect every $\frac{dy}{dx}$ term, factor, and divide …

Common Mistakes

Mistake 1: Treating the $y$ on the right-hand side as a constant

In $xy+y^2=\tan x+y$, the lone $y$ on the right is easy to skip past, but it is still a function of $x$. Why it's wrong: differentiating a $y$-term to $0$ silently drops a $\frac{dy}{dx}$ term and changes the whole equation. Correct approach: every occurrence of $y$, however simple, differentiates to at least $\frac{dy}{dx}$.

Mistake 2: Forgetting the product rule on $xy$ …