Q.Find $\frac{dy}{dx}$, if $y + \sin y = \cos x$.

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.

$\frac{d}{dx}(y) = \frac{dy}{dx}$, and $\frac{d}{dx}(\sin y) = \cos y \cdot \frac{dy}{dx}$ (chain rule).

Right side: $\frac{d}{dx}(\cos x) = -\sin x$.

Step 2: Collect all $\frac{dy}{dx}$ terms:

$\frac{dy}{dx} + \cos y \cdot \frac{dy}{dx} = -\sin x$ …

This problem is a classic example of implicit differentiation. Why can't we just solve for $y$ and differentiate normally? Because the equation $y + \sin y = \cos x$ mixes $y$ and $x$ in a way that cannot be untangled — there's no simple algebraic way to write $y$ as a function of $x$ alone. So we treat $y$ as an unknown function of $x$, and differentiate every term with respect to $x$, using the chain rule whenever we hit a $y$.

The key idea: whenever you differentiate a term involving $y$, you multiply by $\frac{dy}{dx}$ because $y$ itself depends on $x$. This is just the chain rule in action.

Let's work through it step by step.

1. Differentiate both sides with respect to $x$. Left side: $y + \sin y$.
   • The derivative of $y$ with respect to $x$ is $\frac{dy}{dx}$.
   • The derivative of $\sin y$ with respect to $x$ is $\cos y \cdot \frac{dy}{dx}$ (chain rule: derivative of $\sin$ is $\cos$, then multiply by derivative of the inside $y$). So the left side becomes:

Right side: $\cos x$.

• The derivative of $\cos x$ with respect to $x$ is $-\sin x$. So the right side becomes:

2. Write the differentiated equation:

3. Factor out $\frac{dy}{dx}$ from the left side: $\frac{dy}{dx} (1 + \cos y) = -\sin x$ …

Method: Implicit Differentiation of an Equation Mixing $x$ and a Function of $y$

This method applies when $y$ appears not just by itself but inside another function (like $\sin y$, $y^2$, or $e^y$) in an equation that cannot easily be solved to give $y$ explicitly in terms of $x$.

Steps

Step 1: Differentiate every term of the equation with respect to $x$, term by term

Treat $y$ throughout as $y(x)$.

Step 2: Apply the chain rule to any term that is "a function of $y$"

If a term is $h(y)$ for some standard function $h$ (sine, cosine, a power, etc.), its derivative with respect to $x$ is $h'(y) \cdot \dfrac{dy}{dx}$ — never just $h'(y)$ alone.

Step 3: Simplify the right-hand side normally

Differentiate any pure-$x$ terms using the ordinary rules.

Step 4: Collect all $\dfrac{dy}{dx}$ terms together and factor

Every term produced by a $y$-containing part of the equation carries a $\dfrac{dy}{dx}$ factor; group them on one side. …

Common Mistakes

Mistake 1: Dropping the $\frac{dy}{dx}$ factor when differentiating $\sin y$

Why it's wrong: Since $y$ is a function of $x$, differentiating $\sin y$ with respect to $x$ requires the chain rule, producing $\cos y \cdot \frac{dy}{dx}$ — writing just $\cos y$ treats $y$ as though it were the independent variable $x$. Correct approach: every time a trig (or any) function of $y$ is differentiated with respect to $x$, tack on the factor $\frac{dy}{dx}$.

Mistake 2: Trying to cancel $\cos y$ as if it were a common factor …