AP Calculus BC 3.5 Selecting Procedures for Calculating Derivatives Study Notes - New Syllabus
AP Calculus BC 3.5 Selecting Procedures for Calculating Derivatives Study Notes- New syllabus
AP Calculus BC 3.5 Selecting Procedures for Calculating Derivatives Study Notes – AP Calculus BC- per latest AP Calculus BC Syllabus.
LEARNING OBJECTIVE
- This topic is intended to focus on the skill of selecting an appropriate procedure for calculating derivatives.
Key Concepts:
- Selecting Procedures for Calculating Derivatives
Selecting Procedures for Calculating Derivatives
Selecting Procedures for Calculating Derivatives
This topic focuses on the ability to choose the most appropriate method to calculate derivatives, based on the structure of the function. Students should practice recognizing when to apply specific rules and techniques from the derivative toolkit.
Methods to Consider:
- Power Rule: For simple polynomial terms \( x^n \).
- Product Rule: When the function is a product of two differentiable functions.
- Quotient Rule: When the function is a ratio of two differentiable functions.
- Chain Rule: For composite functions, e.g., functions inside another function.
- Implicit Differentiation: When the function is not given explicitly as \( y = f(x) \).
- Inverse Function Rule: For finding derivatives of inverse functions.
- Inverse Trigonometric Functions: For arcsin, arctan, etc.
- Logarithmic Differentiation: For functions involving products, quotients, and powers simultaneously (useful for complex functions like \( y = (x^2+1)^5(x^3+2) \)).
| Rule | Formula | Example |
|---|---|---|
| Power Rule | \( \dfrac{d}{dx}[x^n] = nx^{n-1} \) | \( \dfrac{d}{dx}[x^5] = 5x^4 \) |
| Product Rule | \( (uv)’ = u’v + uv’ \) | \( \dfrac{d}{dx}[x^2\sin x] = 2x\sin x + x^2\cos x \) |
| Quotient Rule | \( \left( \dfrac{u}{v} \right)’ = \dfrac{u’v – uv’}{v^2} \) | \( \dfrac{d}{dx}\Big[\dfrac{\sin x}{x^2}\Big] = \dfrac{x^2\cos x – 2x\sin x}{x^4} \) |
| Chain Rule | \( \dfrac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x) \) | \( \dfrac{d}{dx}[(3x^2+1)^5] = 5(3x^2+1)^4 \cdot 6x \) |
| Implicit Differentiation | Differentiate both sides, treat \( y \) as function of \( x \) | For \( x^2+y^2=25\), \( 2x+2y\dfrac{dy}{dx}=0\) ⇒ \( \dfrac{dy}{dx}=-\dfrac{x}{y} \) |
| Inverse Function Rule | \( (f^{-1})'(a)=\dfrac{1}{f'(b)}, f(b)=a \) | If \( f(x)=x^3+1,a=9 \): \( b=2,f'(x)=3x^2\). So \( (f^{-1})'(9)=\dfrac{1}{12} \) |
| Inverse Trigonometric Functions | \( \dfrac{d}{dx}[\sin^{-1}x]=\dfrac{1}{\sqrt{1-x^2}} \) | \( \dfrac{d}{dx}[\sin^{-1}(3x)]=\dfrac{3}{\sqrt{1-9x^2}} \) |
| Logarithmic Differentiation | Take \( \ln \) on both sides, then differentiate | If \( y=x^x\), \( \ln y = x\ln x \), so \( y’ = x^x(\ln x + 1) \) |
General Strategy:
- Analyze the function structure: Look for sums, products, quotients, or compositions.
- Break it into parts: Apply the correct rule for each part.
- Combine with Chain Rule: Whenever an inner function is involved.
Example:
Differentiate \( y = (3x^2 + 1)^5 \).
▶️ Answer/Explanation
Step 1: Recognize it as a composite function → Apply Chain Rule.
\( y’ = 5(3x^2 + 1)^4 \cdot (6x) = 30x(3x^2 + 1)^4 \).
Answer: \( 30x(3x^2 + 1)^4 \).
Example:
Differentiate \( y = \dfrac{\sin x}{x^2 + 1} \).
▶️ Answer/Explanation
Function is a quotient → Apply Quotient Rule.
\( y’ = \dfrac{(x^2+1)\cos x – (\sin x)(2x)}{(x^2+1)^2} \).
Answer: \( \dfrac{(x^2+1)\cos x – 2x\sin x}{(x^2+1)^2} \).
Example:
Differentiate \( y = x^x \).
▶️ Answer/Explanation
Function has variable in base and exponent → Use Logarithmic Differentiation.
\( \ln y = x\ln x \).
Differentiate: \( \dfrac{1}{y} y’ = \ln x + 1 \).
So \( y’ = x^x (\ln x + 1) \).
\( y’ = x^x (\ln x + 1) \).