Sum and Difference Rule in Differentiation

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. Sum Rule in Derivative
• 2. Difference Rule in Derivative
• 3. Solved Examples

Sum Rule: If $f, g$ and $h$ are differentiable function then:
$\dfrac{d}{dx}\bigg[f(x) + g(x) + h(x) \bigg] = \dfrac{d}{dx}\bigg[f(x)\bigg] + \dfrac{d}{dx}\bigg[g(x)\bigg] + \dfrac{d}{dx}\bigg[h(x)\bigg]$

Difference Rule: If $f, g$ and $h$ are differentiable function then:
$\dfrac{d}{dx}\bigg[f(x) - g(x) - h(x) \bigg] = \dfrac{d}{dx}\bigg[f(x)\bigg] - \dfrac{d}{dx}\bigg[g(x)\bigg] - \dfrac{d}{dx}\bigg[h(x)\bigg]$

Example 1:

Differentiate $f(x) = 5x^4 + 2x^2 - 4x + 5$.

Solution:

$f'(x) = \dfrac{d}{dx}\big( 5x^4 + 2x^2 - 4x + 5 \big)$

$f'(x) = \dfrac{d}{dx}\big( 5x^4\big) + \dfrac{d}{dx}\big(2x^2 \big) - \dfrac{d}{dx}\big(4x \big) + \dfrac{d}{dx}\big(5 \big)$.

$f'(x) = \boxed{20x^3 + 4x - 4}$

Example 2:

Differentiate $f(x) = x^{\pi} - 2x^{\sqrt{2}}$.

Solution:

$f'(x) = \dfrac{d}{dx}\big( x^{\pi} - 2x^{\sqrt{2}} \big)$

$f'(x) = \dfrac{d}{dx}\big( x^{\pi}\big) + \dfrac{d}{dx}\big(2x^{\sqrt{2}} \big)$.

$f'(x) = \boxed{\pi x^{\pi - 1} + 2\sqrt{2}x^{\sqrt{2}-1}}$

Example 3:

Find the equation of the tangent line to $f(x) = 4x - x^2$ at $x = 1$.

Solution:

We know that the equation of a tangent line at $x = a$ is given by:
$\boxed{y = f(a) + f'(a)(x - a)}$

Given $a = 1$ and $f(x) = 4x - x^2$

We can write: $f(1) = 4 - (1)^2 = 3$

Derivative of $f(x)$ is: $f'(x) = 4 - 2x$ then $f'(1) = 4 - 2(1) = 2$

The equation of Tangent is:

$y = 3 + (2)(x - 1) = \boxed{2x + 1}$

Example 4:

The equation of motion of the particle is $s(t) = 2t^2 - 3t + 2$ where $s$ is measured in meters and $t$ is seconds. Find velocity of particle after 1 second.

Solution:

We know that velocity is the derivative of displacement $s(t)$ with respect to time $t$, i.e.,

$v(t) = \dfrac{ds}{dt}$.

Step 1: Differentiate $s(t)$

Differentiating $s(t) = 2t^2 - 3t + 2$ using basic derivative rules:

$v(t) = \dfrac{d}{dt} \big( 2t^2 - 3t + 2 \big)$

Applying derivatives:

$v(t) = 2(2t) - 3(1) + 0$

$v(t) = 4t - 3$.

Step 2: Find $v(1)$

Substituting $t = 1$:

$v(1) = 4(1) - 3$

$v(1) = 4 - 3 = 1$.

Final Answer:

The velocity of the particle after 1 second is: $\boxed{\mathbf{1}}$ m/s.