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Product, Quotient & Chain Rules

Product, Quotient, and Chain Rules

When finding derivatives of complex functions, the basic power rule isn't always enough. You will frequently encounter functions that are multiplied together, divided, or nested inside one another. To differentiate these, we use three fundamental rules of calculus: the Product Rule, the Quotient Rule, and the Chain Rule.

The Product Rule

The Product Rule is used when you need to find the derivative of two functions multiplied together.

Formula: ddx[f(x)g(x)]=f′(x)g(x)+f(x)g′(x)\frac{d}{dx}[f(x)g(x)] = f'(x)g(x) + f(x)g'(x)

In words: The derivative of the first times the second, plus the first times the derivative of the second.

Example: Find ddx(x2⋅ex)\frac{d}{dx}(x^2 \cdot e^x)

  1. Let f(x)=x2f(x) = x^2 and g(x)=exg(x) = e^x.
  2. Find their derivatives: f′(x)=2xf'(x) = 2x and g′(x)=exg'(x) = e^x.
  3. Apply the rule: ddx(x2⋅ex)=(2x)(ex)+(x2)(ex)\frac{d}{dx}(x^2 \cdot e^x) = (2x)(e^x) + (x^2)(e^x)
  4. Factor out the common term to simplify: ex(2x+x2)e^x(2x + x^2)

The Quotient Rule

The Quotient Rule is used when differentiating one function divided by another.

Formula: ddx[f(x)g(x)]=f′(x)g(x)−f(x)g′(x)[g(x)]2\frac{d}{dx}\left[\frac{f(x)}{g(x)}\right] = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2}

In words: "Low D-High minus High D-Low, over the square of what's below."

Example: Find ddx2x+1x2−3\frac{d}{dx}\frac{2x + 1}{x^2 - 3}

  1. Let f(x)=2x+1f(x) = 2x + 1 (high) and g(x)=x2−3g(x) = x^2 - 3 (low).
  2. Find their derivatives: f′(x)=2f'(x) = 2 and g′(x)=2xg'(x) = 2x.
  3. Apply the rule: ddx2x+1x2−3=2(x2−3)−(2x+1)(2x)(x2−3)2\frac{d}{dx}\frac{2x + 1}{x^2 - 3} = \frac{2(x^2 - 3) - (2x + 1)(2x)}{(x^2 - 3)^2}
  4. Expand and simplify the numerator: =2x2−6−(4x2+2x)(x2−3)2=−2x2−2x−6(x2−3)2= \frac{2x^2 - 6 - (4x^2 + 2x)}{(x^2 - 3)^2} = \frac{-2x^2 - 2x - 6}{(x^2 - 3)^2}

The Chain Rule

The Chain Rule is arguably the most important rule in differential calculus. It is used to differentiate composite functions (functions inside other functions).

Formula: ddx[f(g(x))]=f′(g(x))⋅g′(x)\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)

In words: Take the derivative of the outside function (leaving the inside alone), then multiply by the derivative of the inside function.

Example: Find ddxsin⁡(3x2+1)\frac{d}{dx}\sin(3x^2 + 1)

  1. Identify the outside function f(u)=sin⁡(u)f(u) = \sin(u) and the inside function g(x)=3x2+1g(x) = 3x^2 + 1.
  2. The derivative of the outside is cos⁡(u)\cos(u).
  3. The derivative of the inside is 6x6x.
  4. Apply the rule: ddxsin⁡(3x2+1)=cos⁡(3x2+1)⋅6x\frac{d}{dx}\sin(3x^2 + 1) = \cos(3x^2 + 1) \cdot 6x
  5. Rearrange for standard notation: 6xcos⁡(3x2+1)6x \cos(3x^2 + 1)