Q.Find $\frac{dy}{dx}$ in the following: $\cos x \cdot \cos 2x \cdot \cos 3x$

Concept: Chain Rule — differentiate a product by taking the natural log first, then differentiating term-by-term.

Let $y = \cos x \cdot \cos 2x \cdot \cos 3x$.

Take $\log$ on both sides:

Differentiate with respect to $x$:

So:

Substitute back $y$:

When you see a function that is a product of several simpler functions — especially ones like $\cos x$, $\cos 2x$, $\cos 3x$ — the direct product rule would be messy: you’d need to apply it twice, and each term would involve derivatives of cosines with different arguments. There’s a cleaner way.

The chain rule is at the heart here, but we first use a trick: logarithmic differentiation. If $y = f(x)$, taking $\log$ of both sides gives $\log y = \log f(x)$. Differentiating both sides with respect to $x$ uses the chain rule on the left: $\frac{1}{y} \frac{dy}{dx} = \frac{d}{dx} [\log f(x)]$. Then multiply by $y$ to get $\frac{dy}{dx}$. This turns a product into a sum of logs, which is much easier to differentiate.

Let’s apply it.

1. Set up the function.

   Let $y = \cos x \cdot \cos 2x \cdot \cos 3x$.

2. Take the natural logarithm of both sides.

3. Differentiate both sides with respect to $x$. On the left, by the chain rule: $\frac{d}{dx} (\log y) = \frac{1}{y} \frac{dy}{dx}$. On the right, differentiate each term. Remember: $\frac{d}{dx} [\log(\cos kx)] = \frac{1}{\cos kx} \cdot (-\sin kx) \cdot k = -k \tan kx$. So:

4. Solve for $\frac{dy}{dx}$. Multiply both sides by $y$:

5. Substitute back $y$.

Factor the negative sign:

Method: Logarithmic Differentiation for a Product of Several Functions

Use this method whenever $y$ is written as a product of three or more factors (here, three cosines with different arguments) — differentiating such a product directly would need the product rule applied twice and quickly gets unwieldy.

Steps

Step 1: Take the natural logarithm of both sides.

The logarithm of a product turns into a sum of logarithms — this is the whole point of the trick, since sums are far easier to differentiate than products.

Step 2: Differentiate both sides with respect to $x$.

On the left, $y$ is a function of $x$, so by the Chain Rule $\frac{d}{dx}\log y = \frac{1}{y}\frac{dy}{dx}$. On the right, differentiate each $\log(\cos kx)$ term, remembering the Chain Rule brings down a factor of $k$:

Step 3: Solve for $\frac{dy}{dx}$ and substitute $y$ back in.

Multiply the sum of tangent terms by $y$ and replace $y$ with the original product $\cos x\cos 2x\cos 3x$ to get the final answer entirely in terms of $x$. This "log first, differentiate the sum, multiply back by $y$" pattern works for any product of powers or trig factors, no matter how many factors there are.

Common Mistakes

Mistake 1: Forgetting the inner-function factor when differentiating $\log(\cos 2x)$ and $\log(\cos 3x)$.

Why it's wrong: $\frac{d}{dx}\log(\cos 2x)$ is $-2\tan 2x$, not $-\tan 2x$ — the $2$ comes from the Chain Rule applied to the inner argument $2x$. Skipping this factor is the single most common error on this type of problem. Correct approach: always differentiate the argument of each cosine separately and carry that coefficient through.

Mistake 2: Leaving the answer in terms of $y$ instead of substituting back.

Why it's wrong: the question asks for $\frac{dy}{dx}$ as a function of $x$; stopping at $\frac{dy}{dx} = -y(\tan x + 2\tan 2x + 3\tan 3x)$ is incomplete. Correct approach: always replace $y$ with the original expression, $\cos x\cos 2x\cos 3x$, as the final step.

Mistake 3: Sign confusion when combining the three negative tangent terms.

Why it's wrong: since $\frac{d}{dx}\log(\cos kx) = -k\tan kx$ for each term, all three terms carry a minus sign, and factoring out $-1$ at the end is easy to get wrong (e.g. dropping the overall minus sign). Correct approach: write out all three negative terms explicitly before factoring, then double-check the sign of the factored expression.