Product Rule

If the two functions 𝑓(𝑥) and 𝑔(𝑥) are differentiable (i.e. the derivative exist) then the product is differentiable and

• (𝑓·𝑔)′ = 𝑓′𝑔 + 𝑓𝑔′

Example

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find derivative of

• 𝑥²(1 - 𝑥)³

we have:

• 𝑓(𝑥) = 𝑥²
• 𝑔(𝑥) = (1 - 𝑥)³

differentiate those functions:

• 𝑓’(𝑥) = 2𝑥
• 𝑔’(𝑥) = -3(1 - 𝑥)²

put together:

• (𝑓𝑔)′ = 2𝑥(1 - 𝑥)³ - 𝑥²3(1 - 𝑥)²
• (𝑓𝑔)′ = (2𝑥(1 - 𝑥) - 𝑥²3)(1 - 𝑥)²
• (𝑓𝑔)′ = 𝑥(2(1 - 𝑥) - 𝑥3)(1 - 𝑥)²
• (𝑓𝑔)′ = 𝑥(2 - 2𝑥 - 3𝑥)(1 - 𝑥)²
• (𝑓𝑔)′ = 𝑥(2 - 5𝑥)(1 - 𝑥)²

Intuition

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Consider the following function:

• (𝑓·𝑔)(𝑥)

(𝑓·𝑔)(𝑥) is equivalent to the area of a rectangle whose sides are 𝑓(𝑥) and 𝑔(𝑥).

Now imagine increasing 𝑥 by 𝑑𝑥, How much would 𝑑𝑓 increase by?

Thus:

• $d(f\cdot g)=\frac{df}{dx}g+f\frac{dg}{dx}+\frac{df}{dx}\frac{dg}{dx}$
• $d(f\cdot g)=\frac{df}{dx}g+f\frac{dg}{dx}$

Quotient Rule

If the two functions 𝑓(𝑥) and 𝑔(𝑥) are differentiable (i.e. the derivative exist) then the quotient is differentiable and

• (𝑓/𝑔)′ = (𝑓′𝑔 - 𝑓𝑔′) / 𝑔²

Example

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find derivative of

• 𝑥²/ (1 - 𝑥)³

we have:

• 𝑓(𝑥) = 𝑥²
• 𝑔(𝑥) = (1 - 𝑥)³

differentiate those functions:

• 𝑓’(𝑥) = 2𝑥
• 𝑔’(𝑥) = -3(1 - 𝑥)²

put together:

• (𝑓/𝑔)′ = [2𝑥(1 - 𝑥)³ + 𝑥²3(1 - 𝑥)²] / (1 - 𝑥)⁶
• (𝑓/𝑔)′ = [(2𝑥(1 - 𝑥) + 𝑥²3)(1 - 𝑥)²] / (1 - 𝑥)⁶
• (𝑓/𝑔)′ = [𝑥(2(1 - 𝑥) + 𝑥3)(1 - 𝑥)²] / (1 - 𝑥)⁶
• (𝑓/𝑔)′ = [𝑥(2 - 2𝑥 + 3𝑥)(1 - 𝑥)²] / (1 - 𝑥)⁶
• (𝑓/𝑔)′ = 𝑥(2 + 𝑥)(1 - 𝑥)² / (1 - 𝑥)⁶
• (𝑓/𝑔)′ = 𝑥(2 + 𝑥) / (1 - 𝑥)⁴