Download 2.3 Notes Implicit Differentiation and Logarithmic

2.3 Implicit Differentiation and Logarithmic Differentiation
☼ Two ways to find y’ in the equation xy=1, explicitly and IMPLICITLY.
First we will use explicit differentiation.
xy  1
1
y
x
y  x 1
y '  1x 2
y' 
1
x2
Now we will solve the problem using IMPLICIT differentiation.
xy  1
x  y ' y 1  0
xy '   y
y' 
y
x
(*This answer looks a little different than when we did the problem explicitly, but the two answers are equivalent:
1
 
y
1
x
y'        2
x
x
x
y
1
x
(1,1)
(2, ½)
1
1
m 2 
2
4

(-1/2, -2)
Examples:
A)
5 y 2  sin y  x 2 (First we will take the derivative with respect to X!!)
5  2 y  y ' cos y  y '  2 x (Algebraically solve for y’.)
y ' 10 y  cos y   2 x
y' 
2x
10 y  cos y
B)
d2y
2
2
Find
if 4 x  2 y  9
2
dx
8x  4 y  y '  0
4 y  y '  8 x
2x
y' 
y
 y 1  x  y ' 
y ''  2 

y2


2x 

 y  x y 

y ''  2 
y2






2
x
2y  4
y  y
y '' 
 
2
y
 y
2 y 2  4 x2
y '' 
y3
y '' 
9
y3
Proof of
d n
x  n  x n 1 for “rational” exponents:
dx
Let
y  x r where r is any rational #. r 
m
where m and n are integers.
n
y  xr
m
y  xn
n
 m
 y   x n 
 
n
m
y x
d
d m
 y n  
x 
dx
dx  
n  y n1  y '  m  x m1
n
m
n
y  x then y
Note: Since
So…
n x
m
y' 
m
n
 y '  mx m1
mx m1
m
m
nx n
m

m m1 m n 
y'  x
n
m m 1 m  mn
y'  x
n
m m 1
y'  xn
n
 r  x r 1
n 1
 m
xn 
 
n 1
x
m n 1
n
x
mn  m
n
mn m

n
xn
x
m
m
n
Recall:
d
1
ln x  
dx
x
d
1
ln x  
dx
x
d
1
ln u    u ' (chain rule when u is some function of x)
dx
u
d
1
logb x 
dx
 ln b  x
Example:
y
x 2 3 7 x  14
1  x 
2 4
 23

x 7 x  14 
(Recall y  x
ln  y   ln 
 1  x 2 4 


logb y  logb x )
*You can only feed logs positive #s. If


ln y  ln x 2 3 7 x  14  ln 1  x 2 
a  b then a  b .
4
ln y  ln x 2  ln  7 x  14  3  ln 1  x 2 
1
4
1
ln y  2 ln x  ln  7 x  14   4 ln 1  x 2  (Try to make u as close to just x as possible.)
3
*Calculus step:
1
1 1
1
1
 y '  2  
7  4
 2x
y
x 3  7 x  14 
1  x2 
2
7
8x 

y'  y 

2
 x 3  7  x  2  1  x  
*Substitute what y was originally to finish solving for y’.
2
1 x 1  x 2 
x 2  7 x  14  3  2  3  x  2  1  x 
8 x  3x  x  2  


y'


4
1  x2   3x  x  2  1  x 2  3x  x  2  1  x 2  3x  x  2  1  x 2  
1
1
x 2  7 x  14  3  6 x3  12 x 2  6 x  12  x  x 3  24 x 3  48 x 2 


y' 
4
3x  x  2  1  x 2 

1  x2  
1
x 2  3 7  x  2  3  17 x3  36 x 2  7 x  12 

 *derivative is undefined when x=0.
y' 
2
2 4
3
x
x

2
1

x






1

x
  
3
y' 
7 x  17 x3  36 x 2  7 x  12 
3  x  2  1  x
2
3

2 5
x0
Proof: Power rule for all real numbers.
Let r be any real number.
y  xr
y  xr
ln y  ln x r
ln y  ln x
r
ln y  r  ln x
d
d
ln y  r  ln x
dx
dx
1
1
 y'  r
y
x
y
y' r
x
xr
y'  r
x
y '  r  x r 1
Examples:
Find the points at which the graph of
2
4 x 2  y 2  8x  4 y  4  0 has vertical and horizontal tangent lines.
8 x  2 yy ' 8  4 y ' 0  0
2 yy ' 4 y '  8  8 x
yy ' 2 y '  4  4 x
 y  2 y '  4  4 x
y' 
4  4x
y2
Vertical tangent line: y= -2
Horizontal tangent line: x= 1
4 x  8x  0
4  y2  8  4 y  4  0
x2  2x  0
y2  4 y  0
x  x  2  0
x0 x2
 0, 2  2, 2
y  y  4  0
y  0 y  4
1,0 1, 4
2
4 x2  y 2  8x  4 y  4  0