Q.Find $\frac{dy}{dx}$ in the following: $e^x \sin 5x$

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

We have $y = e^x \sin 5x$. This is a product, so first use the Product Rule:

Now apply the Chain Rule to $\sin 5x$: derivative is $\cos 5x \cdot 5 = 5\cos 5x$.

The derivative of $e^x$ is $e^x$.

So: …

The function $e^x \sin 5x$ is a product of two distinct functions: $e^x$ and $\sin 5x$. When you have a product, the Product Rule is your first instinct. But notice that $\sin 5x$ itself is not a simple sine — it’s a composition: $\sin(5x)$. That means the Chain Rule will be needed inside the product.

So the core idea: Product Rule first, then Chain Rule on the second factor.

Let’s walk through it.

1. Identify the two factors.

   Let $u = e^x$ and $v = \sin 5x$. Then $y = u \cdot v$.

2. Differentiate $u$.

   The derivative of $e^x$ is simply $e^x$ itself:

   $\frac{du}{dx} = e^x$.

3. Differentiate $v = \sin 5x$ using the Chain Rule.

   The outer function is $\sin(\cdot)$, whose derivative is $\cos(\cdot)$. The inner function is $5x$, whose derivative is $5$.

   So:

   $\frac{dv}{dx} = \cos(5x) \cdot 5 = 5 \cos 5x$.

   Tip

   A quick check: derivative of $\sin(kx)$ is always $k \cos(kx)$. This pattern saves time.

4. Apply the Product Rule.

   The Product Rule says: $\frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx}$.

   Substitute:

   $\frac{dy}{dx} = e^x \cdot (5 \cos 5x) + \sin 5x \cdot e^x$. …

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

$\frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}.$ …

Common Mistakes

Mistake 1: Forgetting the chain-rule factor of $5$ when differentiating $\sin5x$.

Why it's wrong: $\frac{d}{dx}\sin5x = 5\cos5x$, not $\cos5x$ — because $5x$ is the inner function, whose own derivative ($5$) must be multiplied in. Correct approach: whenever a trig function's argument is not simply $x$, differentiate the argument as an explicit chain-rule step.

Mistake 2: Forgetting to factor out $e^x$ from both terms in the final answer. …