By: Jane Kim, Period 4, 2007 Source: Mr. Wiencek’s Noted from Class.
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Transcript of By: Jane Kim, Period 4, 2007 Source: Mr. Wiencek’s Noted from Class.
By: Jane Kim , Period 4, 2007
Source: Mr. Wiencek’s Noted from Class
The Definition of Derivative
f(x)
(x,0)
(x,f(x))
((x+h),f(x+h))
h
((x+h),0)
Limit Definition of a Derivative
h 0limf ‘(x) =
f(x+h) – f(x)
h
f ‘(x) = limh 0
f(x+h) – f(x)
h
Remember!
• Don’t forget to write out the Limit Definition of Derivative
• Remember to write everytime limh 0
Example
f(x) = 5x + 3
F(x+h) = 5(x+h) + 3
= 5x + 5h + 3
5x + 5h + 3 – (5x + 3)
h
Example continued
5x + 5h + 3 – 5x – 3
h
5h
h
= 5Answer: 5
Power Rule, Slopes of Tangent Lines
• f ‘(x) F Prime of x
• y’ y prime
• dy
dx
• d
dx
dy dx
Derivatives with respect to x
Common Powers
x = x
5
x = x
1x = x
1
x3= x
LN Yx = x
x = x43
x
x
x
x
x
x
1/2
1/5
-1
- 3
4
4/3
Power Rule
y = x
y’ = x*– 1
Example
y = 3x - x + 2
y’ = 2(3)x – 1(-1) + 0(2)
= 6x + (-1)x + 0x
= 6x – 1
2
2 - 1 1 - 1 0 - 1
0 -1
Answer: = 6x – 1
dy
dxdy
dx
Remember!
• Derivatives = Slope
Example
y = 2x x = 0,1,3,-4
f(0) = 4(0) = 0
f(1) = 4(1) = 4
f(3) = 4(3) = 12
f(-4) = 4(-4) = -16
dy
dx = 4x
2
Graphs & Using the Derivative to find Slope
Tangent Line
Slope = m
Normal Line
Slope = 1m
Example
y = 2x + 3Find the equation ofa) The tangent at 1b) The normal at 1
y = 2x + 3y’= 6x + 3y(1) = 6(1) + 3
= 6 + 3 = 9
3
3
2
(1, ?)
**Derivatives = Slope Slope = 9
Example continued
To find y: plug x = 1 back into the original equation, y = 2x + 3
y = 2(1) + 3(1)
= 2 + 3
= 5
so (1,5)
3
3
(1, ?)
Example continued
Tangent equation:
y – y = m(x – x )
y – 5 = 9(x – 1)
y – 5 = 9x – 9
y = 9x – 4
Normal equation:
y – y = - 1/9(x – x )
y – 5 = - 1/9(x – 1)
y – 5 = -1/9x + 1/9
y = -1/9x + 46/9
1 1 1 1
THE END