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

Differentiate $2x+3y=\sin y$ implicitly, treating $y$ as a function of $x$:

Collect the derivative terms:

The relation $2x+3y=\sin y$ can't be solved neatly for $y$, so we differentiate both sides with respect to $x$, remembering every $y$-term carries a factor $\dfrac{dy}{dx}$.

Differentiate term by term

So

Solve for the derivative

Move the $\dfrac{dy}{dx}$ terms together:

Since $\cos y\le 1<3$, the factor $3-\cos y$ is always positive, so we can divide safely:

The derivative is negative everywhere; multiplying top and bottom by $-1$ gives the equivalent form $\dfrac{2}{\cos y-3}$.

Method: Implicit Differentiation When y Appears on Both Sides of the Equation

Use this method when $y$ shows up in more than one term of the equation, including inside a function like $\sin y$ or $\cos y$ — this requires collecting the $\dfrac{dy}{dx}$ terms together before you can solve for the derivative.

Steps

Step 1: Differentiate both sides term by term, applying the chain rule to every $y$-term

Every occurrence of $y$ — whether it's $y$ by itself or tucked inside another function — produces a factor of $\dfrac{dy}{dx}$ when differentiated. For $\sin y$: $\dfrac{d}{dx}\sin y = \cos y\cdot\dfrac{dy}{dx}$.

Step 2: Move every term containing $\dfrac{dy}{dx}$ to one side of the equation, and everything else to the other

After Step 1, $\dfrac{dy}{dx}$ typically appears in more than one term — some coming from the left side of the original equation, some from the right. Collect them all together algebraically before proceeding.

Step 3: Factor $\dfrac{dy}{dx}$ out of the collected terms

Once every $\dfrac{dy}{dx}$-term is on the same side, factor it out as a common factor, leaving a single bracket multiplying $\dfrac{dy}{dx}$.

Step 4 (Applying to this problem): Divide by the bracketed coefficient to isolate $\dfrac{dy}{dx}$

Since the coefficient often still contains $y$ (not just $x$), the final answer is left in terms of both $x$ and $y$ — this is expected and correct for implicit differentiation, not a sign anything went wrong.

Common Mistakes

Mistake 1: Differentiating $\sin y$ as $\cos y$ instead of $\cos y\cdot\dfrac{dy}{dx}$

Why it's wrong: $\sin y$ is a composite function of $x$ (since $y$ depends on $x$), so its derivative needs the chain rule just as much as any other $y$-term — treating it like $\sin x$ and forgetting the extra factor is a very common slip precisely because the function itself doesn't visually "look different" from the explicit case. Correct approach: mentally substitute $y=y(x)$ before differentiating any trig/exponential function of $y$, so the chain-rule factor is never forgotten.

Mistake 2: Moving the $\dfrac{dy}{dx}$ terms to the wrong side, causing a sign error

Why it's wrong: the equation $2+3\dfrac{dy}{dx}=\cos y\,\dfrac{dy}{dx}$ has $\dfrac{dy}{dx}$-terms on both sides; subtracting incorrectly (e.g. moving the $\cos y\,\dfrac{dy}{dx}$ term without flipping its sign) leaves the wrong coefficient in the final bracket. Correct approach: rewrite the equation so all $\dfrac{dy}{dx}$-terms sit on one designated side, doing the subtraction one term at a time and tracking each sign explicitly.

Mistake 3: Treating the final answer (which still contains $y$) as incomplete or "not fully solved"

Why it's wrong: some students try to further substitute or eliminate $y$ from the answer, but since the original equation cannot be solved for $y$ explicitly in the first place, an answer like $\dfrac{dy}{dx}=\dfrac{-2}{3-\cos y}$ in terms of both $x$ (implicitly) and $y$ is the correct final form. Correct approach: recognize that a derivative expressed in terms of both variables is the expected, complete answer for implicit differentiation.