Derivative of Product of Three or More Functions

If a function is written as a product of three or more differentiable functions:

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

Then its derivative is obtained by differentiating one function at a time and keeping others unchanged:

[f’(x)][=u’(x)\cdot v(x)\cdot w(x)][+u(x)\cdot v’(x)\cdot w(x)][+u(x)\cdot v(x)\cdot w’(x)+\dots]

Differentiate one factor at a time
Add all such terms

Example 1

Differentiate:
[f(x)][=x\cdot(x+1)\cdot(x^{2}+2)]

Step-by-Step Solution

1. Let
   [u=x], [v=(x+1)], [w=(x^{2}+2)]
2. Differentiate each:
   [u’=1], [v’=1], [w’=2x]
3. Apply the product rule for 3 factors:
   [f’(x)][=u’vw+uv’w+uvw’]
4. Substitute:
   [f’(x)][=1\cdot(x+1)\cdot(x^{2}+2)][+x\cdot1\cdot(x^{2}+2)][+x\cdot(x+1)\cdot2x]
5. Simplify:
   = [(x+1)(x^{2}+2)][+x(x^{2}+2)][+2x^{2}(x+1)]
6. Expand and combine:
   = [(x^{3}+x^{2}+2x+2)][+(x^{3}+2x)][+(2x^{3}+2x^{2})]
   = [4x^{3}+3x^{2}+4x+2]

Conclusion:
[f’(x)][=4x^{3}+3x^{2}+4x+2]

Example 2

Differentiate:
[f(x)][=(x-2)(x^{2}+3x)(2x+5)]

Step-by-Step Solution

1. Let
   [u=(x-2)], [v=(x^{2}+3x)], [w=(2x+5)]
2. Derivatives:
   [u’=1], [v’=2x+3], [w’=2]
3. Apply rule:
   [f’(x)][=u’vw+uv’w+uvw’]
4. Substitute:
   [f’(x)][=1\cdot(x^{2}+3x)(2x+5)][+(x-2)(2x+3)(2x+5)][+(x-2)(x^{2}+3x)\cdot2]
5. Expand (optional for exam unless required)
   Final answer may be left in factored form.

Conclusion:
[f’(x)][=(x^{2}+3x)(2x+5)][+(x-2)(2x+3)(2x+5)][+2(x-2)(x^{2}+3x)]