Q.Find $\frac{dy}{dx}$ in the following: $\cos x^3 \cdot \sin^2 (x^5)$

Concept: Chain Rule — differentiate outer functions, then multiply by the derivative of the inner function.

Let $y = \cos(x^3) \cdot \sin^2(x^5)$. Use the product rule first:

Now apply the chain rule to each term:

• $\frac{d}{dx}\cos(x^3) = -\sin(x^3) \cdot 3x^2$
• $\frac{d}{dx}\sin^2(x^5) = 2\sin(x^5) \cdot \cos(x^5) \cdot 5x^4$ (chain rule on $\sin$, then on $x^5$)

Substitute back: …

The problem asks for $\frac{dy}{dx}$ where $y = \cos(x^3) \cdot \sin^2(x^5)$. This is a product of two functions, each of which is a composition of simpler functions. The key is to see the structure clearly: we have an outer function (like $\cos u$ or $v^2$) wrapped around an inner function ($x^3$ or $x^5$). The Chain Rule tells us to differentiate the outer function first, then multiply by the derivative of the inner function. And because it's a product, we also need the Product Rule.

Let’s break it down step by step.

1. Identify the structure. Write $y = f(x) \cdot g(x)$, where $f(x) = \cos(x^3)$ and $g(x) = \sin^2(x^5)$. The Product Rule says:

2. Differentiate $f(x) = \cos(x^3)$. Here the outer function is $\cos u$ and the inner function is $u = x^3$. The derivative of $\cos u$ is $-\sin u$, so:

3. Differentiate $g(x) = \sin^2(x^5)$. This is a composition of three functions: $(\sin(x^5))^2$. Think of it as $h^2$ where $h = \sin(x^5)$. The derivative of $h^2$ is $2h \cdot h'$, so:

Now $\frac{d}{dx}[\sin(x^5)]$ is another Chain Rule: derivative of $\sin v$ is $\cos v$, with $v = x^5$, so:

Putting it together:

Method: Product Rule for Two Composite Functions

Use this method when you must differentiate a product $y = f(x)\cdot g(x)$ where BOTH factors are themselves composite (chain-rule) functions, not simple polynomials.

Steps

Step 1: Split the product into its two factors

Write $y = f(x)\cdot g(x)$ and identify each factor clearly before differentiating anything.

Step 2: Differentiate each factor on its own, using the chain rule as many times as that factor needs

Treat each factor as a self-contained mini-problem. A factor like $\sin^2(x^5)$ is itself a composition of three functions — squaring, then sine, then $x^5$ — so it needs the chain rule applied twice on its own, before you even get to the product rule:

Step 3: Apply the product rule using the fully-differentiated factors from Step 2

$\frac{d}{dx}\big[f(x)g(x)\big] = f'(x)g(x) + f(x)g'(x).$ …

Common Mistakes

Mistake 1: Differentiating $\sin^2(x^5)$ as if it were a single chain-rule layer

Why it's wrong: $\sin^2(x^5)$ has three layers (square, sine, $x^5$), so it needs the chain rule applied twice — once for the square, once for the sine of $x^5$. Writing $\frac{d}{dx}\sin^2(x^5) = 2\sin(x^5)$ stops after only the outer square and forgets to multiply by $\frac{d}{dx}\sin(x^5) = 5x^4\cos(x^5)$. Correct approach: work the composite factor as its own mini chain-rule problem, layer by layer, before plugging it into the product rule.

Mistake 2: Mixing up the product rule's two terms …