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

Idea: $y = x\cos x$ is a product of two functions, so use the product rule: $(uv)' = u'v + uv'$.

Let $u=x$ and $v=\cos x$. Then $u'=1$ and $v'=-\sin x$, so

The function $y = x\cdot\cos x$ is a product of two functions of $x$: namely $x$ and $\cos x$. You cannot just differentiate each factor and multiply — that would wrongly give $-\sin x$. Products need the product rule.

Set up

Take $u = x$ and $v = \cos x$.

Differentiate each part

Apply the rule

Method: The Product Rule (with Chain Rule on Each Factor)

When two functions of $x$ are multiplied together, neither the sum rule nor differentiating each factor separately and multiplying works — the product rule is required.

Steps

Step 1: Identify the two factors $u(x)$ and $v(x)$ being multiplied

Step 2: Differentiate each factor separately

If either factor is itself composite, apply the chain rule to it individually at this stage.

Step 3: Combine using the product rule

Step 4: Factor out any common terms to simplify

Applying to this problem: for $y=x\cos x$, take $u=x$ ($u'=1$) and $v=\cos x$ ($v'=-\sin x$); the product rule gives $\frac{dy}{dx}=\cos x - x\sin x$.

Common Mistakes

Mistake 1: Differentiating each factor separately and multiplying the results, instead of using the product rule.

Why it's wrong: $\frac{d}{dx}(x)\cdot\frac{d}{dx}(\cos x) = 1\cdot(-\sin x) = -\sin x$ is NOT the derivative of $x\cos x$ — differentiation does not distribute over multiplication. Correct approach: always apply $u'v+uv'$, never $u'v'$.

Mistake 2: Sign error on $\frac{d}{dx}\cos x$.

Why it's wrong: $\cos x$ differentiates to $-\sin x$, and writing $+\sin x$ here would flip the sign of the second term in the final answer. Correct approach: keep $\frac{d}{dx}\cos x=-\sin x$ memorised alongside $\frac{d}{dx}\sin x=+\cos x$ to avoid mixing them up.