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

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:

This gives:

Isolate $\frac{dy}{dx}$:

This is a straightforward implicit differentiation problem. The equation $2x + 3y = \sin x$ already has $y$ mixed with $x$, but we can still differentiate both sides with respect to $x$ — the key is to remember that $y$ is a function of $x$, so when we differentiate $3y$, we use the chain rule.

Let’s walk through it.

1. Differentiate every term on both sides with respect to $x$.

   • The derivative of $2x$ is $2$.
   • The derivative of $3y$ is $3 \cdot \frac{dy}{dx}$ (because $y$ depends on $x$, we multiply by $\frac{dy}{dx}$).
   • The derivative of $\sin x$ is $\cos x$.

   So we get:

2. Isolate $\frac{dy}{dx}$. Subtract $2$ from both sides:

3. Divide by 3:

That’s it. No extra manipulation needed because the equation is linear in $y$.

Method: Implicit Differentiation for an Equation Linear in y

Use this method whenever $y$ appears in an equation only to the first power — never squared, never inside a trig/exponential function, and never multiplied by $x$. This is the simplest case of implicit differentiation.

Steps

Step 1: Differentiate every term on both sides of the equation with respect to $x$

Treat $y$ as an unknown function of $x$, not as an independent variable — this matters even though $y$ appears "by itself" here.

Step 2: Apply the chain rule to the $y$-term

Because $y=y(x)$, differentiating a term like $ky$ (a constant times $y$) gives $k\cdot\dfrac{dy}{dx}$, not just $k$. This single extra factor is the whole difference between implicit and explicit differentiation.

Step 3: Differentiate every $x$-only term using ordinary rules

Terms containing only $x$ (no $y$) differentiate exactly as in single-variable calculus — no extra chain-rule factor is needed for these.

Step 4 (Applying to this problem): Isolate $\dfrac{dy}{dx}$ algebraically

Since $y$ appears only linearly, $\dfrac{dy}{dx}$ shows up in exactly one term after differentiating. Move every other term to the opposite side of the equation, then divide by the constant coefficient in front of $\dfrac{dy}{dx}$ to solve for it directly.

Common Mistakes

Mistake 1: Differentiating $3y$ as $3$ instead of $3\dfrac{dy}{dx}$

Why it's wrong: because $y$ looks like a "plain variable," it's tempting to differentiate it the same way as a constant times $x$ — but $y$ is secretly a function of $x$, so the chain rule adds the factor $\dfrac{dy}{dx}$ that must not be dropped. Correct approach: whenever a term contains $y$, always ask "am I differentiating with respect to $x$ or $y$?" — since the answer is $x$, every $y$ picks up a $\dfrac{dy}{dx}$.

Mistake 2: Forgetting to isolate $\dfrac{dy}{dx}$ by dividing by its full coefficient

Why it's wrong: after differentiating, the equation reads $2+3\dfrac{dy}{dx}=\cos x$; leaving the answer as $3\dfrac{dy}{dx}=\cos x-2$ without the final division by $3$ is an incomplete answer, not a wrong method — but it costs marks since the question asks for $\dfrac{dy}{dx}$ itself. Correct approach: always finish by dividing through by the coefficient so $\dfrac{dy}{dx}$ stands alone.