Differentiation rules (product, quotient, chain) Knowledge-Based Study

Study notes

1. Overview of Operations on Derivatives

   Basic differentiation rules, such as the power rule and constant multiple rule, allow us to differentiate linear combinations of functions term by term. However, when functions are multiplied, divided, or composed with one another, we cannot simply differentiate each component independently.

   To find the rates of change for these combined functions, specific operational rules are required:

   • Product Rule: for differentiating products of two differentiable functions.
   • Quotient Rule: for differentiating ratios of two differentiable functions.
   • Chain Rule: for differentiating composite functions.

   Mastering these rules enables the differentiation of complex algebraic, trigonometric, and transcendental functions.

   Differentiating Products of Functions

   If $u(x)$ and $v(x)$ are both differentiable functions, their product $f(x) = u(x) \cdot v(x)$ is also differentiable. The derivative is not simply the product of their individual derivatives.

   The Product Rule states:

   $$\frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x)$$

   In words: differentiate the first function and multiply by the second, then add the first function multiplied by the derivative of the second.

   Worked Example

   Differentiate $f(x) = (3x^2 - 1)(2x + 5)$:

   1. Identify components: $u(x) = 3x^2 - 1$ and $v(x) = 2x + 5$.
   2. Find individual derivatives: $u'(x) = 6x$ and $v'(x) = 2$.
   3. Apply the formula: $f'(x) = (6x)(2x + 5) + (3x^2 - 1)(2)$.
   4. Expand and simplify: $f'(x) = 12x^2 + 30x + 6x^2 - 2 = 18x^2 + 30x - 2$.

   Differentiating Ratios of Functions

   When a function is expressed as a quotient $f(x) = \frac{u(x)}{v(x)}$ where $v(x) \neq 0$, its derivative is determined using the Quotient Rule:

   $$\frac{d}{dx}\left[\frac{u(x)}{v(x)}\right] = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}$$

   A common mnemonic is "low d-high minus high d-low over the square of what's below".

   Worked Example

   Find the derivative of $g(x) = \frac{x^2 + 3}{2x - 1}$ for $x \neq \frac{1}{2}$:

   1. Identify components: $u(x) = x^2 + 3$ and $v(x) = 2x - 1$.
   2. Compute derivatives: $u'(x) = 2x$ and $v'(x) = 2$.
   3. Apply the quotient formula:
      $$g'(x) = \frac{(2x)(2x - 1) - (x^2 + 3)(2)}{(2x - 1)^2}$$
   4. Expand the numerator: $4x^2 - 2x - (2x^2 + 6) = 2x^2 - 2x - 6$.
   5. Express final result: $$g'(x) = \frac{2x^2 - 2x - 6}{(2x - 1)^2}$$

   Differentiating Composite Functions

   The Chain Rule is used when one function is nested inside another, represented as $y = f(g(x))$. If $g$ is differentiable at $x$ and $f$ is differentiable at $g(x)$, then the derivative of the composite function is:

   $$\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)$$

   Using Leibniz notation with $y = f(u)$ and $u = g(x)$:

   $$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$$

   General Power Rule

   A frequent application of the chain rule is differentiating powers of functions, $[g(x)]^n$:

   $$\frac{d}{dx}[g(x)]^n = n[g(x)]^{n-1} \cdot g'(x)$$

   Worked Example

   Find $h'(x)$ for $h(x) = (4x^3 - 5x)^6$:

   1. Outer function: $f(u) = u^6$, with $f'(u) = 6u^5$.
   2. Inner function: $u = g(x) = 4x^3 - 5x$, with $g'(x) = 12x^2 - 5$.
   3. Combine using chain rule:
      $$h'(x) = 6(4x^3 - 5x)^5 \cdot (12x^2 - 5)$$

   Composite Applications

   Many calculus problems require combining two or more rules in a single problem, such as using the chain rule inside a product or quotient.

   Worked Example: Product and Chain Rules Together

   Differentiate $y = x^2 \sqrt{2x + 1}$:

   1. Rewrite with rational exponents: $y = x^2 (2x + 1)^{1/2}$.
   2. Treat as a product $u(x) \cdot v(x)$, where $u(x) = x^2$ and $v(x) = (2x + 1)^{1/2}$.
   3. Differentiate $u(x)$: $u'(x) = 2x$.
   4. Differentiate $v(x)$ using the chain rule:
      $$v'(x) = \frac{1}{2}(2x + 1)^{-1/2} \cdot 2 = (2x + 1)^{-1/2}$$
   5. Assemble using the product rule:
      $$y' = 2x(2x + 1)^{1/2} + x^2(2x + 1)^{-1/2}$$
   6. Factor out common terms with the lowest power $(2x + 1)^{-1/2}$:
      $$y' = (2x + 1)^{-1/2} \left[ 2x(2x + 1) + x^2 \right] = \frac{4x^2 + 2x + x^2}{\sqrt{2x + 1}} = \frac{5x^2 + 2x}{\sqrt{2x + 1}}$$