Derivative of Product of Two Functions

1. Concept Overview

When a function is the product of two simpler functions, we do not differentiate each term separately!
Instead, we apply the Product Rule.

If:

[f(x)=u(x)\cdot v(x)]

Then derivative:

[\dfrac{d}{dx}(u\cdot v) = u\dfrac{dv}{dx} + v\dfrac{du}{dx}]

In words:
First × derivative of second + Second × derivative of first

A helpful mnemonic:

“UV rule: First D-second + Second D-first”

---

2. Step by Step Examples

Example 1

Differentiate [f(x)=x\sin x].

Solution
1. Let [u=x], [v=\sin x]
2. [\dfrac{du}{dx}=1], [\dfrac{dv}{dx}=\cos x]
3. Apply Product Rule:
[f'(x)=u\dfrac{dv}{dx} + v\dfrac{du}{dx}]
[f'(x)=x\cos x + \sin x]

✔ Final Answer:
[f'(x)=x\cos x + \sin x]

---

Example 2

Differentiate [f(x)=x^{2}\cdot e^{x}].

Solution
1. [u=x^{2}], [v=e^{x}]
2. [\dfrac{du}{dx}=2x], [\dfrac{dv}{dx}=e^{x}]
3. Apply rule:
[f'(x)][=x^{2}e^{x} + e^{x}(2x)]
[f'(x)][=e^{x}(x^{2}+2x)]

✔ Result:
[f'(x)][=e^{x}(x^{2}+2x)]

---

Example 3

Differentiate [f(x)=x\cdot \ln x], where [x>0].

Solution
1. [u=x], [v=\ln x]
2. [\dfrac{du}{dx}=1], [\dfrac{dv}{dx}=\dfrac{1}{x}]
3. [f'(x)][=x\cdot\dfrac{1}{x} + \ln x\cdot 1]
[f'(x)=1 + \ln x]

✔ Final Answer
[f'(x)=\ln x + 1], [x>0]

---

3. Important Notes

Example: [7x^{3}] → [7\cdot 3x^{2} = 21x^{2}] ✔ No product rule needed

---

4. Conceptual Questions with Solutions (15)

1. Why can’t we simply differentiate u and v separately in a product?

Because differentiation does not distribute over multiplication. We must account for the change in both functions simultaneously using the Product Rule.

2. In [\dfrac{d}{dx}(uv)], what does the term [u\dfrac{dv}{dx}] represent?

Change in product caused by change in v while u remains fixed.

3. In [\dfrac{d}{dx}(uv)], what does the term [v\dfrac{du}{dx}] represent?

Change in product caused by change in u while v remains fixed.

4. What does “First D-Second” mean?

Multiply first function by derivative of second.

5. What does “Second D-First” mean?

Multiply second function by derivative of first.

6. When differentiating [kx], do we apply Product Rule?

No. k is a constant → use simple rule: [\dfrac{d}{dx}(kx)=k].

7. Which rule is more powerful: product rule or sum rule?

Product rule — because it includes the effect of two simultaneously changing functions.

8. Can product rule be extended to more than two factors?

Yes, but then apply the rule repeatedly between pairs of functions.

9. Do we need product rule in [\sin x\cdot\cos x]?

Yes — both are variable functions.

10. In what cases does [\dfrac{d}{dx}(uv)=0]?

If the product is constant, like [u=\dfrac{1}{v}], derivative may become zero.

11. Does the product rule work for piecewise functions?

Yes, as long as the functions are differentiable in the region of interest.

12. Can product rule be used for polynomials times trigonometric functions?

Yes — very common case.

13. Do we apply product rule in [5\ln x]?

No. 5 is a constant multiplier.

14. Who discovered the product rule?

It emerged from the work of Newton and Leibniz during the development of calculus.

15. What is the minimum condition for product rule?

Both u and v must be differentiable at that point.

---

5. FAQ / Common Misconceptions

1. Is [\dfrac{d}{dx}(uv)=u’v’] correct?

❌ No. That misses the interactions between u and v.

2. Does product rule apply to constant × function?

No. Only when both parts vary.

3. Can we reverse the order?

Yes: [u\dfrac{dv}{dx} + v\dfrac{du}{dx}] is commutative in sum.

4. Is product rule same as chain rule?

No. Product rule is for multiplication; chain rule is for composition.

5. Do we need parentheses after differentiating?

Yes, especially in simplification.

6. Should we differentiate inside multiplication first?

Identify u and v first → then differentiate.

7. Does product always get bigger?

No — derivative may be zero or negative.

8. If u or v is zero, does derivative vanish?

Not necessarily; the other function’s derivative still contributes.

9. Do we use product rule in [\ln(x^{2}+1)]?

No — that would require **chain rule**, not product rule.

10. Can we change factor order?

Yes — multiplication is commutative.

---

6. Practice Questions — Step by Step Solutions

---

Question 1

Differentiate [x^{2}\sin x].

Solution
1. [u=x^{2}], [v=\sin x]
2. [u’=2x], [v’=\cos x]
3. [f’=x^{2}\cos x + 2x\sin x]

Conclusion
[\dfrac{d}{dx}(x^{2}\sin x)][=x^{2}\cos x+2x\sin x]

---

Question 2

Differentiate [x^{3}e^{x}] (treat as product rule; no chain rule).

Solution
1. [u=x^{3}], [v=e^{x}]
2. [u’=3x^{2}], [v’=e^{x}]
3. [f’=x^{3}e^{x}+3x^{2}e^{x}]
[f’][=e^{x}(x^{3}+3x^{2})]

Conclusion
[\dfrac{d}{dx}(x^{3}e^{x})][=e^{x}(x^{3}+3x^{2})]

---

Question 3

Differentiate [x\cos x].

Solution
1. [u=x], [v=\cos x]
2. [u’=1], [v’=-\sin x]
3. [f’=x(-\sin x)+\cos x]
[f’=\cos x – x\sin x]

Conclusion
[\dfrac{d}{dx}(x\cos x)][=\cos x – x\sin x]

---

Question 4

Differentiate [x^{2}\ln x], [x>0].

Solution
1. [u=x^{2}], [v=\ln x]
2. [u’=2x], [v’=\dfrac{1}{x}]
3. [f’][=x^{2}\cdot\dfrac{1}{x} + 2x\cdot\ln x]
[f’=x + 2x\ln x]

Conclusion
[\dfrac{d}{dx}(x^{2}\ln x)][=x(1+2\ln x)], [x>0]

---

Question 5

Differentiate [\sin x \cdot \cos x].

Solution
1. [u=\sin x], [v=\cos x]
2. [u’=\cos x], [v’=-\sin x]
3. [f’=\sin x(-\sin x) + \cos x(\cos x)]
[f’][=\cos^{2}x – \sin^{2}x]

Conclusion
[\dfrac{d}{dx}(\sin x\cos x)][=\cos^{2}x – \sin^{2}x]

---

Question 6

Differentiate [x\cdot\sec x].

Solution
1. [u=x], [v=\sec x]
2. [u’=1], [v’=\sec x\tan x]
3. [f’=x\sec x\tan x + \sec x]
[f’=\sec x(x\tan x + 1)]

Conclusion
[\dfrac{d}{dx}(x\sec x)][=\sec x(x\tan x+1)]

---

Question 7

Differentiate [x^{4}\cdot \tan x].

Solution
1. [u=x^{4}], [v=\tan x]
2. [u’=4x^{3}], [v’=\sec^{2}x]
3. [f’][=x^{4}\sec^{2}x + 4x^{3}\tan x]

Conclusion
[\dfrac{d}{dx}(x^{4}\tan x)][=x^{4}\sec^{2}x + 4x^{3}\tan x]

---

Question 8

Differentiate [x\cdot \ln x], [x>0].

Solution
(Already shown — repeated for drilling)
[f'(x)=\ln x + 1]

---

Question 9

Differentiate [xe^{x}], simple template.

Solution
1. [u=x], [v=e^{x}]
2. [u’=1], [v’=e^{x}]
3. [f’=xe^{x}+e^{x}]
[f’=e^{x}(x+1)]

Conclusion
[\dfrac{d}{dx}(xe^{x})][=e^{x}(x+1)]

---

Question 10

Differentiate [x^{2}\cot x].

Solution
1. [u=x^{2}], [v=\cot x]
2. [u’=2x], [v’=-\csc^{2}x]
3. [f’=x^{2}(-\csc^{2}x) + 2x\cot x]
[f’][=-x^{2}\csc^{2}x + 2x\cot x]

Conclusion
[\dfrac{d}{dx}(x^{2}\cot x)][= -x^{2}\csc^{2}x + 2x\cot x]