Q.If $\cos y = x \cos (a+y)$, with $\cos a \neq \pm 1$, prove that $\frac{dy}{dx} = \frac{\cos^2 (a+y)}{\sin a}$.

Concept: Implicit Differentiation — differentiate both sides with respect to $x$, treating $y$ as a function of $x$.

Step 1: Differentiate $\cos y = x \cos(a+y)$ implicitly:

Step 2: Rearrange to isolate $\frac{dy}{dx}$:

Step 3: From the original equation, $x = \frac{\cos y}{\cos(a+y)}$. Substitute:

$\frac{dy}{dx} \left[ \frac{\cos y}{\cos(a+y)} \sin(a+y) - \sin y \right] = \cos(a+y)$ …

The core idea here is implicit differentiation. The equation $\cos y = x \cos(a+y)$ ties $x$ and $y$ together in a way that we cannot (and need not) solve for $y$ explicitly. Instead, we differentiate both sides with respect to $x$, remembering that $y$ is a function of $x$, so every time we hit a $y$, we multiply by $\frac{dy}{dx}$ (the chain rule). Then we isolate $\frac{dy}{dx}$ and use the original equation to simplify.

Let’s work through it step by step.

1. Differentiate both sides with respect to $x$. Left side: $\frac{d}{dx}[\cos y] = -\sin y \cdot \frac{dy}{dx}$. Right side: $\frac{d}{dx}[x \cos(a+y)]$. This is a product: $x$ times $\cos(a+y)$.
   • Derivative of $x$ is $1$, so the first term: $1 \cdot \cos(a+y) = \cos(a+y)$.
   • Derivative of $\cos(a+y)$ is $-\sin(a+y) \cdot \frac{dy}{dx}$ (chain rule again, since $a$ is constant). Multiply by $x$: $x \cdot [-\sin(a+y) \frac{dy}{dx}] = -x \sin(a+y) \frac{dy}{dx}$. So the derivative of the right side is:

Putting it together:

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

Factor out $\frac{dy}{dx}$:

3. Solve for $\frac{dy}{dx}$:

This is a valid expression, but it still contains $x$ and $\sin y$. We want it purely in terms of $a$ and $y$. That’s where the original equation comes in.

4. Use the original relation to eliminate $x$. From $\cos y = x \cos(a+y)$, we have:

Substitute this into the denominator:

Combine into a single fraction:

Notice the numerator: $-\sin y \cos(a+y) + \cos y \sin(a+y)$. This is exactly $\sin(a+y - y)$? Let’s check: …

Method: Implicit Differentiation, Then Use the Original Equation to Simplify

For a relation like $\cos y = x\cos(a+y)$ that ties $x$ and $y$ together (with $a$ a constant), differentiate both sides with respect to $x$ (every $y$-term picks up a factor of $\dfrac{dy}{dx}$), collect the $\dfrac{dy}{dx}$ terms, and — crucially — use the original equation again to eliminate any leftover $x$ from the final expression.

Steps

Step 1: Differentiate both sides implicitly with respect to $x$

Apply the product rule to the right side (since it's $x$ times a function of $y$), and the chain rule wherever $y$ appears.

Step 2: Collect all $\dfrac{dy}{dx}$ terms on one side and factor

Step 3: Solve for $\dfrac{dy}{dx}$

At this point the expression may still contain $x$.

Step 4: Substitute the original relation back in to eliminate $x$ …

Common Mistakes

Mistake 1: Forgetting the product rule on the right side $x\cos(a+y)$.

Why it's wrong: the right side is a product of $x$ and $\cos(a+y)$, both of which effectively vary — skipping the product rule and only differentiating $\cos(a+y)$ misses the $\cos(a+y)$ term that comes from differentiating $x$ itself. Correct approach: apply the product rule fully, treating $x$ and $\cos(a+y)$ as the two factors.

Mistake 2: Stopping once $\dfrac{dy}{dx}$ is solved for, without eliminating $x$ using the original relation. …