Selecting Procedures for
Calculating Derivatives
Master the art of choosing the right differentiation technique for any function: power rule, product rule, quotient rule, chain rule, and beyond.
AP Calculus BC
What Is a Derivative?
A derivative measures the instantaneous rate of change of a function. Geometrically, it gives the slope of the tangent line at any point on a curve.
The challenge on the AP exam is not just computing derivatives: it is recognizing which rule to apply quickly and accurately.
The Core Differentiation Rules
Power Rule
\( \frac{d}{dx}[x^n] = nx^{n-1} \)
Product Rule
\( \frac{d}{dx}[fg] = f'g + fg' \)
Quotient Rule
\( \frac{d}{dx}\!\left[\frac{f}{g}\right] = \frac{f'g - fg'}{g^2} \)
Chain Rule
\( \frac{d}{dx}[f(g(x))] = f'(g(x))\cdot g'(x) \)
The Decision Flowchart
Always scan for composition first: if any part is nested inside another function, the chain rule is required, possibly layered with other rules.
The Power Rule in Depth
\( f(x) = x^5 \)
\( f'(x) = 5x^4 \)
\( f(x) = \sqrt{x} = x^{1/2} \)
\( f'(x) = \tfrac{1}{2}x^{-1/2} \)
\( f(x) = \frac{1}{x^3} = x^{-3} \)
\( f'(x) = -3x^{-4} \)
Key insight: Before reaching for the quotient rule, rewrite radicals and reciprocals as power functions. This often simplifies the work dramatically.
Product vs. Quotient Rule
| Product Rule | Quotient Rule | |
|---|---|---|
| Form | \( f(x) \cdot g(x) \) | \( \frac{f(x)}{g(x)} \) |
| Formula | \( f'g + fg' \) | \( \frac{f'g - fg'}{g^2} \) |
| Mnemonic | "first d-second + second d-first" | "lo d-hi minus hi d-lo over lo-lo" |
| Sign | Addition (+) | Subtraction (-) |
| Tip | Simpler, fewer errors | Rewrite as product with \( g^{-1} \) when possible |
Unpacking the Chain Rule
1
Identify layers: What is the outer function? What is the inner function?
2
Differentiate the outer as if the inner were a simple variable.
3
Multiply by the inner derivative \( g'(x) \).
4
Repeat if there are more nested layers (multi-chain).
Essential Derivative Formulas
Trig
\( (\sin x)' = \cos x \)
\( (\cos x)' = -\sin x \)
\( (\tan x)' = \sec^2 x \)
\( (\cos x)' = -\sin x \)
\( (\tan x)' = \sec^2 x \)
Exp / Log
\( (e^x)' = e^x \)
\( (a^x)' = a^x \ln a \)
\( (\ln x)' = \frac{1}{x} \)
\( (a^x)' = a^x \ln a \)
\( (\ln x)' = \frac{1}{x} \)
Inverse Trig
\( (\arcsin x)' = \frac{1}{\sqrt{1-x^2}} \)
\( (\arctan x)' = \frac{1}{1+x^2} \)
\( (\arctan x)' = \frac{1}{1+x^2} \)
Step-by-Step Selection Procedure
1
Simplify first: Expand products, combine like terms, rewrite radicals and fractions as powers.
2
Identify the outermost operation: Is it a sum, product, quotient, or composition?
3
Match the rule: Sum → differentiate term-by-term. Product → product rule. Quotient → quotient rule (or rewrite). Composition → chain rule.
4
Apply recursively: Each sub-expression may itself require its own rule. Work from outside inward.
5
Simplify the result: Factor, combine fractions, and reduce for a clean final answer.
Example 1: Power + Sum
1
Identify: This is a polynomial: a sum of power functions. No product, quotient, or composition.
2
Rule: Differentiate term-by-term using the power rule.
3
Compute: \( 3 \cdot 4x^3 - 5 \cdot 2x + 7 \cdot 1 - 0 \)
Example 2: Product + Chain
1
Outermost operation: A product of \( f = x^2 \) and \( g = \sin(3x) \).
2
Product rule: \( f'g + fg' = 2x\sin(3x) + x^2 \cdot [\sin(3x)]' \)
3
Chain rule on \( \sin(3x) \): outer = \( \sin(u) \), inner = \( 3x \). So \( [\sin(3x)]' = \cos(3x) \cdot 3 \).
Example 3: Quotient or Rewrite?
Method A: Quotient Rule
\( \frac{e^x \cdot x^3 - e^x \cdot 3x^2}{x^6} = \frac{e^x(x-3)}{x^4} \)
Method B: Rewrite
\( e^x \cdot x^{-3} \), then product rule:
\( e^x \cdot x^{-3} + e^x(-3x^{-4}) = \frac{e^x(x-3)}{x^4} \)
\( e^x \cdot x^{-3} + e^x(-3x^{-4}) = \frac{e^x(x-3)}{x^4} \)
Example 4: Multi-Layer Chain Rule
1
Layer 1 (outermost): \( \ln(u) \) where \( u = \cos(x^2) \). Derivative: \( \frac{1}{\cos(x^2)} \).
2
Layer 2: \( \cos(v) \) where \( v = x^2 \). Derivative: \( -\sin(x^2) \).
3
Layer 3 (innermost): \( x^2 \). Derivative: \( 2x \).
4
Multiply all layers: \( \frac{1}{\cos(x^2)} \cdot (-\sin(x^2)) \cdot 2x \).
Advanced Techniques
Implicit Differentiation
When \( y \) is not isolated: differentiate both sides with respect to \( x \), treating \( y \) as \( y(x) \). Every time you differentiate a \( y \)-term, attach \( \frac{dy}{dx} \) via the chain rule. Then solve for \( \frac{dy}{dx} \).
Logarithmic Differentiation
For \( y = f(x)^{g(x)} \) (variable base and exponent): take \( \ln \) of both sides, differentiate implicitly, then solve. Essential for forms like \( x^x \) or \( (\sin x)^{x} \).
Common Mistakes to Avoid
✕ Forgetting the Chain Rule
Writing \( \frac{d}{dx}[\sin(3x)] = \cos(3x) \) without the factor of \( 3 \). Always multiply by the inner derivative.
✕ Product Rule Sign Error
Confusing the product rule (\( + \)) with the quotient rule (\( - \)). Remember: product uses addition, quotient uses subtraction.
✕ Incorrect Quotient Rule Order
Swapping \( f'g - fg' \) to \( fg' - f'g \). The correct numerator is \( f'g - fg' \): differentiate the numerator first, multiply by the denominator, then subtract.
✕ Not Simplifying First
Using the quotient rule on \( \frac{x^5}{x^2} \) instead of simplifying to \( x^3 \) first. Always look for algebra shortcuts before differentiating.
Summary of Derivative Procedures
| Function Form | Rule to Use | Key Action |
|---|---|---|
| \( cx^n \) | Power Rule | Bring down exponent, reduce by 1 |
| \( f + g \) or \( f - g \) | Sum / Difference | Differentiate each term separately |
| \( f \cdot g \) | Product Rule | \( f'g + fg' \) |
| \( f / g \) | Quotient Rule | \( \frac{f'g - fg'}{g^2} \) |
| \( f(g(x)) \) | Chain Rule | Outer' \( \times \) inner' |
| Equation with \( y \) | Implicit | Differentiate both sides, solve for \( \frac{dy}{dx} \) |
| \( f(x)^{g(x)} \) | Log Differentiation | Take ln, differentiate, solve |
Question 1 of 5
Which rule is the primary rule needed to differentiate \( f(x) = (2x+1)^5 \)?
The function \( (2x+1)^5 \) is a composite function: the outer function is \( u^5 \) and the inner function is \( u = 2x+1 \). The chain rule gives \( 5(2x+1)^4 \cdot 2 = 10(2x+1)^4 \). The power rule alone would miss the inner derivative factor of 2.
Question 2 of 5
What is \( \frac{d}{dx}\!\left[x \cdot e^x\right] \)?
This is a product of \( f = x \) and \( g = e^x \). Applying the product rule: \( f'g + fg' = 1 \cdot e^x + x \cdot e^x = e^x + xe^x = e^x(1+x) \). Choice D incorrectly applies the power rule to \( e^x \), which is not valid since the base is not \( x \).
Question 3 of 5
Find \( \frac{d}{dx}\!\left[\frac{\sin x}{x}\right] \).
This is a quotient with \( f = \sin x \) and \( g = x \). The quotient rule gives \( \frac{f'g - fg'}{g^2} = \frac{\cos x \cdot x - \sin x \cdot 1}{x^2} = \frac{x\cos x - \sin x}{x^2} \). Choice C reverses the subtraction order (a very common mistake). Choice B ignores the quotient rule entirely.
Question 4 of 5
What is \( \frac{d}{dx}\!\left[e^{\sin x}\right] \)?
This is a composite function: outer = \( e^u \), inner = \( u = \sin x \). The chain rule gives \( e^{\sin x} \cdot \cos x \). Choice A forgets the chain rule. Choice B incorrectly changes \( \sin x \) to \( \cos x \) in the exponent.
Question 5 of 5
A student needs to differentiate \( h(x) = \frac{x^4}{x^2 + 1} \). Which approach is most efficient?
The function is a quotient of two expressions that cannot be simplified (since \( x^2 + 1 \) does not factor to cancel with \( x^4 \)). The quotient rule directly applies, or you can rewrite as \( x^4(x^2+1)^{-1} \) and use the product + chain rule. Logarithmic differentiation works but is overkill. You cannot simply cancel \( x^2 \) from the denominator \( x^2 + 1 \).