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

The key idea is implicit differentiation — since $y$ is not isolated, we differentiate both sides with respect to $x$, treating $y$ as a function of $x$.

Step 1: Differentiate term by term:

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

Step 3: Collect $\frac{dy}{dx}$ terms on one side:

Step 4: Factor and solve:

The equation $ax + by^2 = \cos y$ mixes $x$ and $y$ in a way we cannot solve for $y$ cleanly. That is exactly when implicit differentiation shines. Instead of isolating $y$ first, we differentiate both sides with respect to $x$, remembering that $y$ is a function of $x$ — so every time we hit a $y$, we apply the chain rule and multiply by $\frac{dy}{dx}$.

Let’s go step by step.

1. Differentiate $ax$

   The derivative of $ax$ with respect to $x$ is simply $a$.

2. Differentiate $by^2$

   Here $y$ is a function of $x$, so by the chain rule:

3. Differentiate $\cos y$ Again, $y$ is inside the cosine, so chain rule gives:

4. Put it together Differentiating both sides of $ax + by^2 = \cos y$ yields:

5. Collect the $\frac{dy}{dx}$ terms Bring the term with $\frac{dy}{dx}$ from the right side to the left:

6. Factor out $\frac{dy}{dx}$

7. Solve for $\frac{dy}{dx}$ Provided $2by + \sin y \neq 0$, we get:

Method: Implicit Differentiation with a Nonlinear y-Term and a Trig-of-y Term Together

Use this method when the equation mixes more than one type of $y$-dependence — for instance a squared term ($y^2$) and a trigonometric term ($\cos y$) both appearing, each requiring its own combination of the chain rule (and, for $y^2$, also the power rule).

Steps

Step 1: Differentiate the polynomial-in-$y$ term using the power rule plus the chain rule

For a term like $by^2$, first apply the ordinary power rule as if $y$ were $x$ (giving $2y$), then multiply by $\dfrac{dy}{dx}$ because $y$ is secretly a function of $x$:

Step 2: Differentiate the trig-of-$y$ term using its own derivative rule plus the chain rule

For $\cos y$, apply the ordinary derivative of cosine (giving $-\sin y$), then multiply by $\dfrac{dy}{dx}$ for the same reason:

Step 3: Differentiate any pure-$x$ or parameter terms normally, with no extra factor

Terms like $ax$ (where $a$ is just a constant parameter) differentiate exactly as in ordinary single-variable calculus, giving simply $a$ — parameters are treated the same as any other constant.

Step 4 (Applying to this problem): Collect every $\dfrac{dy}{dx}$ term, factor, and divide

Gather every term carrying a $\dfrac{dy}{dx}$ factor (from both Steps 1 and 2, and from either side of the original equation) onto one side, factor $\dfrac{dy}{dx}$ out of the resulting bracket, and divide by that bracket to isolate $\dfrac{dy}{dx}$ — the same collecting-and-factoring procedure used for any implicit differentiation problem, just with more terms to track here.

Common Mistakes

Mistake 1: Applying the power rule to $by^2$ but forgetting the chain-rule factor $\dfrac{dy}{dx}$

Why it's wrong: writing $\dfrac{d}{dx}(by^2)=2by$ correctly does the power-rule part but stops short — since $y$ is a function of $x$, an extra $\dfrac{dy}{dx}$ must be multiplied on, giving $2by\,\dfrac{dy}{dx}$. Correct approach: treat every $y$-term as requiring the power rule (or trig/exp rule) followed by a chain-rule multiplication, never one without the other.

Mistake 2: Losing the minus sign when differentiating $\cos y$

Why it's wrong: $\dfrac{d}{dx}\cos y = -\sin y\,\dfrac{dy}{dx}$, and the negative sign is easy to drop when attention is on remembering the chain-rule factor — students may correctly write $\sin y\,\dfrac{dy}{dx}$ but omit the leading minus. Correct approach: write out the full unmodified derivative rule for cosine ($-\sin$) before attaching the chain-rule factor, rather than combining both steps mentally.

Mistake 3: Confusing the constant parameters $a$ and $b$ with variables to be solved for

Why it's wrong: since $a$ and $b$ are fixed parameters (not the unknowns), differentiating $ax$ should simply give $a$ — some students mistakenly try to isolate or manipulate $a$ and $b$ as if they were additional unknowns alongside $\dfrac{dy}{dx}$. Correct approach: treat $a$ and $b$ exactly like ordinary numeric constants throughout the differentiation and algebra.