AP Calculus AB / Unit 2: Differentiation: Definition and Fundamental Properties
2.6 Derivative Rules: Constant, Sum, Difference, and Constant Multiple
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Notes
Rule 1: constants
$\displaystyle \frac{d}{dx}\,c = 0$a constant never changes, so its slope is 0
Why constants disappear
Rule 2: constant multiple
$\displaystyle \frac{d}{dx}\,[\,c\,f(x)\,] = c\,f'(x)$keep the number, take the derivative of the rest
$\displaystyle \frac{d}{dx}\,7x^3 = 7 \cdot 3x^2 = 21x^2$example
Rule 3: sum and difference
$\displaystyle \frac{d}{dx}\,[\,f(x) \pm g(x)\,] = f'(x) \pm g'(x)$take each term's derivative on its own
Example 1
- $\displaystyle f(x) = 5x^3 - 4x^2 + 6$three terms
- $\displaystyle 5 \cdot 3x^2 = 15x^2$first term
- $\displaystyle 4 \cdot 2x = 8x$second term
- $\displaystyle 6 \rightarrow 0$the constant
- $\displaystyle f'(x) = 15x^2 - 8x$put it together
Common mistake
$\displaystyle \frac{d}{dx}(x^2+3) = 2x+3$
$\displaystyle \frac{d}{dx}(x^2+3) = 2x$
the +3 goes away
Example 2: simplify first
- $\displaystyle f(x) = \frac{x^3 + 2x}{x}$there's x on the bottom
- $\displaystyle f(x) = x^2 + 2$divide each term by x first
- $\displaystyle f'(x) = 2x$now it's easy
Derivative rules
- Constant: $\frac{d}{dx}c = 0$
- Constant multiple: keep the number in front
- Sum and difference: one term at a time
- Simplify first when the bottom is one term
Practice
From the video
If $f(x) = 5x^3 - 4x^2 + 6$, find $f'(2)$.
It's forty four. f prime of x is fifteen x squared minus eight x. Fifteen times four is sixty. Minus sixteen is forty four.
Find $\displaystyle \frac{d}{dx}\,(7x^3)$.
Keep the 7 in front: $7 \cdot 3x^2 = 21x^2$.
Find $\displaystyle \frac{d}{dx}\,(x^2 + 3)$.
The derivative of the constant $3$ is $0$, so the answer is $2x$.
Find $\displaystyle \frac{d}{dx}\,(5x^3 - 4x^2 + 6)$.
$5 \cdot 3x^2 = 15x^2$, $4 \cdot 2x = 8x$, and the constant 6 goes to 0.
Find $\displaystyle \frac{d}{dx}\,\frac{x^3 + 2x}{x}$.
Simplify first: $\displaystyle \frac{x^3+2x}{x} = x^2 + 2$. The derivative is $2x$.
If $g(x) = 4f(x)$ and $f'(3) = 5$, find $g'(3)$.
Constant multiple rule: $g'(3) = 4f'(3) = 4 \cdot 5 = 20$.
If $h(x) = f(x) - g(x)$, $f'(1) = 6$ and $g'(1) = -2$, find $h'(1)$.
Difference rule: $h'(1) = f'(1) - g'(1) = 6 - (-2) = 8$.
Find $\displaystyle \frac{d}{dx}\left(\frac{3}{x} - 2\sqrt{x}\right)$.
$\displaystyle \frac{3}{x} = 3x^{-1}$ gives $-3x^{-2}$. $\displaystyle -2\sqrt{x} = -2x^{1/2}$ gives $\displaystyle -x^{-1/2} = -\frac{1}{\sqrt{x}}$.
Free response
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Let $f(x) = 2x^3 - 6x + 1$.
(a) Find $f'(x)$.
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$f'(x) = 6x^2 - 6$
Give yourself the points you earned:
(b) Find all values of $x$ where the tangent line to the graph of $f$ is horizontal.
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A horizontal tangent has slope $0$: $6x^2 - 6 = 0$, so $x^2 = 1$ and $x = 1$ or $x = -1$.
Give yourself the points you earned:
(c) Find the slope of the tangent line at $x = 2$.
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$f'(2) = 6(4) - 6 = 18$
Give yourself the points you earned:
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