Product Rule for Derivatives

Neetesh Kumar | February 08, 2025 $\space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space$ Share this Page on:

• 1. Definition of Product Rule
• 2. Explanation & Intuition
• 3. Solved Examples
• 4. Conclusion

Product Rule:
If $u(x)$ and $v(x)$ are differentiable functions, then the derivative of their product is given by:

$\boxed{\dfrac{d}{dx} \bigg[ u(x). v(x) \bigg] = u(x). \dfrac{d}{dx} \bigg[v(x)\bigg] + v(x). \dfrac{d}{dx} \bigg[u(x)\bigg]}$

This means that when differentiating a product of two functions, we take the derivative of the first function and multiply it by the second function, then add the product of the first function and the derivative of the second function.

Example 1: Differentiating a Polynomial and Exponential Function

Differentiate $f(x) = x^2 e^x$.

Solution:

Using the product rule where $u(x) = x^2$ and $v(x) = e^x$:

$f'(x) = x^2 \dfrac{d}{dx} e^x + e^x \dfrac{d}{dx} x^2$

Since $\dfrac{d}{dx} (e^x) = e^x$ and $\dfrac{d}{dx} (x^2) = 2x$, we get:

$f'(x) = x^2 e^x + e^x (2x)$

$f'(x) = e^x (x^2 + 2x)$

Thus, the final answer is: $\boxed{e^x (x^2 + 2x)}$.

Example 2: Differentiating a Trigonometric Function

Differentiate $f(x) = x \sin x$.

Solution:

Using the product rule, let $u(x) = x$ and $v(x) = \sin x$:

$f'(x) = x .\dfrac{d}{dx} (\sin x) + \sin x. \dfrac{d}{dx} (x)$

Since $\dfrac{d}{dx} (\sin x) = \cos x$ and $\dfrac{d}{dx} (x) = 1$, we get:

$f'(x) = x. \cos x + \sin x. (1)$

$f'(x) = x .\cos x + \sin x$

Thus, the final answer is: $\boxed{x \cos x + \sin x}$.

Example 3: Finding the Rate of Change of Kinetic Energy

The kinetic energy of an object is given by $K = \dfrac{1}{2} m v^2$, where $m$ is mass (constant) and $v$ is velocity. Find the rate of change of kinetic energy with respect to time.

Solution:

Since $K = \dfrac{1}{2} m v^2$, differentiate both sides with respect to $t$:

$\dfrac{dK}{dt} = \dfrac{1}{2} .m. \dfrac{d}{dt} (v^2)$

Using the product rule, let $u = v$ and $v = v$, then:

$\dfrac{d}{dt} (v^2) = 2 v. \dfrac{dv}{dt}$

So,

$\dfrac{dK}{dt} = \frac{1}{2} m (2 v \dfrac{dv}{dt})$ $= m v \dfrac{dv}{dt}$

Thus, the rate of change of kinetic energy is: $\boxed{m v \dfrac{dv}{dt}}$.