Physics and Mathematics
Derivative of Sum and Difference of Two Functions
1. Statement of the Concept
The derivative of the sum (or difference) of two differentiable functions is equal to the sum (or difference) of their derivatives.
Mathematically:
- For sum:
[\dfrac{d}{dx}(u(x) + v(x))] [= \dfrac{du}{dx} + \dfrac{dv}{dx}] - For difference:
[\dfrac{d}{dx}(u(x) – v(x)) = \dfrac{du}{dx} – \dfrac{dv}{dx}]
2. Clear Explanation
Suppose [y = u + v], where [u] and [v] are differentiable functions of [x].
Then small changes in [y], called [\Delta y], are the sum of small changes in [u] and [v]:
[\Delta y = \Delta u + \Delta v]
Divide both sides by [\Delta x]:
[\dfrac{\Delta y}{\Delta x}] [= \dfrac{\Delta u}{\Delta x} + \dfrac{\Delta v}{\Delta x}]
Taking the limit as [\Delta x \to 0]:
[\dfrac{dy}{dx} = \dfrac{du}{dx} + \dfrac{dv}{dx}]
Similarly, for the difference:
[\dfrac{d}{dx}(u – v)] [= \dfrac{du}{dx} – \dfrac{dv}{dx}]
Thus, differentiation distributes over addition and subtraction.
3. Key Features
| Feature | Description |
|---|---|
| Linearity | Differentiation is a linear operation |
| No extra rules | Simply differentiate each term individually |
| Works only for sum/difference | Multiplication uses product rule — not this rule |
| Simplifies complex expressions | Breaks into smaller functions |
4. Important Formulas to Remember
| Function | Differentiation Rule | Result |
|---|---|---|
| [u(x) + v(x)] | [\dfrac{d}{dx}(u+v)] | [\dfrac{du}{dx} + \dfrac{dv}{dx}] |
| [u(x) – v(x)] | [\dfrac{d}{dx}(u-v)] | [\dfrac{du}{dx} – \dfrac{dv}{dx}] |
5. Conceptual Questions with Solutions
1. Why can we differentiate each term separately?
Because differentiation is a **linear** operation. The limit process applies independently to each function.
2. Does the rule apply to subtraction?
Yes. [\dfrac{d}{dx}(u-v)] [= \dfrac{du}{dx} – \dfrac{dv}{dx}]
3. Can we extend this to more than two functions?
Yes — any finite sum or difference: [\dfrac{d}{dx}(u+v+w)] [= \dfrac{du}{dx} + \dfrac{dv}{dx} + \dfrac{dw}{dx}]
4. Do constants affect differentiation?
No — constants remain the same and differentiate as usual: [\dfrac{d}{dx}(u + 5)] [= \dfrac{du}{dx}]
5. Can the rule be used with trigonometric or exponential functions?
Yes — the rule applies to **all** differentiable functions.
6. FAQ / Common Misconceptions
1. Is this rule same as the Product Rule?
No — product rule is different: [\dfrac{d}{dx}(uv)] [= u\dfrac{dv}{dx} + v\dfrac{du}{dx}]
2. Can we differentiate terms with different variables?
No — all terms must be functions of the **same variable**.
3. Can we rearrange the order before differentiation?
Yes — addition/subtraction are commutative.
4. Does derivative of a difference increase rate of change?
Not necessarily — it depends on the functions involved.
5. What if one function is not differentiable?
Then the whole sum/difference cannot be differentiated by this rule.
7. Practice Questions (with Step-by-Step Solutions)
Question 1
Find [\dfrac{d}{dx}(x^2 + 3x – 7)].
Step-by-Step Solution:
- Differentiate each term:
[\dfrac{d}{dx}(x^2)=2x],; [\dfrac{d}{dx}(3x)=3],; [\dfrac{d}{dx}(-7)=0] - Add results:
[2x + 3]
Conclusion: [\dfrac{d}{dx}(x^2+3x-7)] [= 2x+3]
Question 2
Find [\dfrac{d}{dx}(\sin x + e^x)].
Solution:
- [\dfrac{d}{dx}(\sin x)=\cos x]
- [\dfrac{d}{dx}(e^x)=e^x]
- Add them:
[\cos x + e^x]
Question 3.
Differentiate [\ln x – \cos x].
Step-by-Step Solution:
- Identify the two functions to differentiate separately:
- [u(x) = \ln x]
- [v(x) = \cos x]
- Differentiate each function using standard formulas:
- [\dfrac{d}{dx}(\ln x) = \dfrac{1}{x}]
- [\dfrac{d}{dx}(\cos x) = -\sin x]
- Apply the difference rule: derivative of [u – v] is [u’ – v’]:
[\dfrac{d}{dx}(\ln x – \cos x)] [= \dfrac{1}{x} – (-\sin x)] - Simplify the signs:
[\dfrac{1}{x} + \sin x]
Conclusion:
[\dfrac{d}{dx}(\ln x – \cos x)] [= \dfrac{1}{x} + \sin x].
Question 4.
Differentiate [e^x + \tan x – x].
Step-by-Step Solution:
- Break into three parts:
- [u(x)=e^x]
- [v(x)=\tan x]
- [w(x)=x]
- Differentiate each part:
- [\dfrac{d}{dx}(e^x)=e^x]
- [\dfrac{d}{dx}(\tan x)=\sec^2 x]
- [\dfrac{d}{dx}(x)=1]
- Use linearity (sum/difference rule):
[\dfrac{d}{dx}(e^x + \tan x – x)] [= e^x + \sec^2 x – 1]
Conclusion:
[\dfrac{d}{dx}(e^x + \tan x – x)] [= e^x + \sec^2 x – 1].
Question 5.
Differentiate [x^3 – 5x + \dfrac{1}{x}].
Step-by-Step Solution:
- Rewrite terms clearly and identify rules:
- [x^3] → power rule with [n=3]
- [-5x] → constant multiple rule
- [\dfrac{1}{x}] → rewrite as [x^{-1}] to use power rule
- Differentiate each term:
- [\dfrac{d}{dx}(x^3) = 3x^2]
- [\dfrac{d}{dx}(-5x) = -5]
- [\dfrac{d}{dx}(x^{-1})] [= -1\cdot x^{-2}] [= -x^{-2}] , which is [ -\dfrac{1}{x^2} ]
- Combine the results using sum/difference rule:
[3x^2 – 5 – \dfrac{1}{x^2}
Conclusion:
[\dfrac{d}{dx}\Big(x^3 – 5x + \dfrac{1}{x}\Big)] [= 3x^2 – 5 – \dfrac{1}{x^2}].
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