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

The key idea is implicit differentiation — since $y$ is not isolated, differentiate both sides with respect to $x$, treating $y$ as a function of $x$, then solve for $\frac{dy}{dx}$.

Step 1: Differentiate $\sin^2 y$ using the chain rule:

Step 2: Differentiate $\cos(xy)$ using the chain rule and product rule:

Step 3: The derivative of the constant $\kappa$ is $0$. So: …

We are given $\sin^2 y + \cos(xy) = \kappa$, where $\kappa$ is a constant. The equation mixes $x$ and $y$ in a way that cannot be solved for $y$ in elementary terms — so we use implicit differentiation.

The core idea: treat $y$ as an unknown function of $x$, i.e. $y = y(x)$. Then every time we differentiate a term containing $y$, we apply the chain rule: the derivative of $f(y)$ with respect to $x$ is $f'(y) \cdot \frac{dy}{dx}$. The constant $\kappa$ differentiates to zero.

Let’s work through it step by step.

1. Differentiate $\sin^2 y$ Write $\sin^2 y = (\sin y)^2$. By the chain rule:

So the first term contributes $2\sin y \cos y \cdot \frac{dy}{dx}$.

2. Differentiate $\cos(xy)$ Here $xy$ is a product of $x$ and $y(x)$, so we need the product rule inside the chain rule. Let $u = xy$. Then $\frac{d}{dx}\cos u = -\sin u \cdot \frac{du}{dx}$. Now $\frac{du}{dx} = \frac{d}{dx}(x y) = 1 \cdot y + x \cdot \frac{dy}{dx}$ (product rule). So

3. Differentiate the constant $\kappa$

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

4. Assemble the differentiated equation

   Putting it all together:

5. Collect terms with $\frac{dy}{dx}$ Expand the second term:

Group the $\frac{dy}{dx}$ terms:

$\left( 2\sin y \cos y - x\sin(xy) \right) \frac{dy}{dx} - y\sin(xy) = 0.$ …

Method: Implicit Differentiation Through a Trig Function of a Product

Use this when $y$ (or a product like $xy$) sits inside a trigonometric function — the chain rule then has to be applied twice: once for the outer trig function, once for what's inside it.

Steps

Step 1: Differentiate a trig function of $y$ alone using the chain rule once

For $f(y)$ where $f$ is a trig function, $\frac{d}{dx}f(y)=f'(y)\cdot\frac{dy}{dx}$. For $\sin^2 y = (\sin y)^2$, this means

Step 2: For a trig function of a product like $xy$, apply the chain rule to the outer function first

Do not stop here — $\frac{d}{dx}(xy)$ itself is not yet computed.

Step 3: Now apply the product rule to what's inside

Substituting back gives the full derivative of the outer trig function — two layers of the chain/product rule stacked together.

Step 4: Differentiate the constant, collect, and solve …

Common Mistakes

Mistake 1: Forgetting the product rule inside the chain rule for $\cos(xy)$

A frequent slip is writing $\frac{d}{dx}\cos(xy)=-\sin(xy)\cdot y$, stopping after applying the outer chain rule but never differentiating the product $xy$ itself. Why it's wrong: the argument of $\cos$ is $xy$, not just $y$, so its own derivative — $y+x\frac{dy}{dx}$ via the product rule — must multiply the $-\sin(xy)$ factor. Correct approach: always ask "what is the derivative of what's inside the trig function?" as a separate sub-step before finishing the chain rule.

Mistake 2: Dropping the $\frac{dy}{dx}$ factor on $\sin^2 y$ …