Download Derivatives

By: Susana Cardona & Demetri
Cheatham
© Cardona & Cheatham 2011
 Slope
 Six
of a tangent line
different techniques: Chain rule, product
rule, Quotient rule, E.T.A, Implicit
differentiation and Logs.
Chain Rule
 Bring exponent down in front of the variable, if it’s a
coefficient multiply exponent. Then subtract one from
the exponent and go back in and take a derivative.

f ( x)  ax n
f ( x)  anx n1
Example

f ( x)  6 x  4 x
3
4
2
3

f ( x)  18 x  16 x

f ( x)  (5 x  1)3
f ( x)  3(5 x  1)2 (5)
Try Me
 f ( x)  ( x  7 x)
3
9
Solution
 f ( x)  9( x  7 x) (3x  7)
3
8
2
Product Rule
 First write the problem times derivative of
the second problem plus write the second
problem times the derivative of the first
problem.
 FDS+SDF
Example
 f ( x)  ( x  3) (3x  1)
2
4
2
3
4

f ( x)  ( x  3) (4)(3x  1) (3)  (3x  1) (2)( x  3)(1)
2
3
3
7
(5
x

1)
(2
x

4)
 f ( x) 
2
3
3
6
2
3
7
2
4

f ( x) (5x 1) (7)(2 x  4) (6 x )  (2 x  4) (3)(5x 1) (10 x)
Try Me
f ( x)  (3x  1) (1  2 x)
4
3
Solution
 f ( x) 
4
3
(3x  1) (3)(1  2 x) (2)  (1  2 x) (4)(3x  1) (3)
4
3

Write the bottom times the derivative of
the top minus write the top times the
derivative of the bottom over the bottom
squared
BDT  TBD
2
B
Quotient Rule
f
( x) 
(5 x  1)
4
(2 x  1)
3
f ( x)  (2x 1) (3)(5x 1) (5)  (5x 1) (4)(2 x 1) (2)
4
2
3
(2 x  1)
8
Example
3
(3
x

5
x
)
◦ f ( x) 
2
4
(6 x  2 x )
2
Try Me
2

f ( x) 
(6 x2  2 x 4 )(2)(3x 2  5x)(6 x  5)  (3x 2  5x)2 (6 x2  2 x 4 )(12 x  8x3 )
(6 x2  2 x4 )2
Solution

Bring down exponent, multiply
coefficient if there’s one, and write the
trig and the angle times the derivative of
the trig times the derivative of the angle
1.
d
(sin 2 3 x)
dx
2sin 3x cos 3 x 3


E
T
A
2.
d
(cos3 (sin x))
dx
2
3cos (sin x)  ( sin(sin x))  (cos x)

d
3x
e
dx

3x
e (3)
NATURAL LOG


1 over the angle times the derivative of the angle
ya
x
dy x
 a ln a
dx
EXAMPLE
y2
x
ln y  x ln 2
1 dy
 x(0)  ln 2(1)
y dx
1 dy
 ln 2
y dx
dy
 2 x ln 2
dx
TRY ME
yx
ln x
SOLUTION
ln y  ln x ln x
1 dy
1
1
 ln x( )  ln x( )
y dx
x
x
dy
ln x 2 ln x
x (
)
dx
x
Implicit
• Is almost the same as a chain rule but it includes
x and y and the x’s and y’s can be separated
•
x2  y 2  1
dy
2x  2 y
0
dx
dy 2 x

dx 2 y
dy  x

dx
y
Example
•
3x3  4 y 2  5
dy
9x  8 y
0
dx
dy
8y
 9 x 2
dx
dy
9 x 2

dx
8y
2
Try Me
3x  2 xy  5 y  1
2
2
Solution
6 x  2 x(1)
dy
dy
 2 y (1)  10 y
0
dx
dx
dy
(2 x  10 y )  2 y  6 x
dx
dy
2 y  6x

dx 2 x  10 y
dy
y  3x

dx  x  5 y
d
4
2
 (tan ( x  2 x  1))
dx
4 tan ( x  2 x  1)  sec ( x  2 x  1)  (2 x  2)
3
2
2
2
y  2x x 1
2
OR
y  (2 x)( x  1)
2
1
2
1

2
1
2
1 2
2
y '  (2 x)( )( x  1) (2 x)  ( x  1) (2)
2
2
y
3
(5 x  1)
4
y '  6(5 x  1) (5)