M53 Lec2.3 Higher Order Derivatives Implicit Differentiation Linear Approximation

Higher Order Derivatives
Implicit Differentiation
Local Linear Approximation and Differentials
Mathematics 53
Institute of Mathematics (UP Diliman)
Institute of Mathematics (UP Diliman) Higher Order, Implicit Differentiation, Differentials Mathematics 53 1 / 35
For today
1 Higher Order Derivatives
2 Implicit Differentiation
3 Local Linear Approximation and Differentials
For today
If f 0 is differentiable,
If f 0 is differentiable, then the derivative of f 0 is called the second derivative of f

and is denoted f 00 .

We can continue to obtain the third derivative, f 000 ,

We can continue to obtain the third derivative, f 000 , the fourth derivative, f (4) ,

and other higher derivatives.

Definition
The n-th derivative of the function f , denoted f (n) , is the derivative of the
(n − 1)-th derivative of f ,

Definition
The n-th derivative of the function f , denoted f (n) , is the derivative of the
(n − 1)-th derivative of f , that is,
f (n−1) ( x + ∆x ) − f (n−1) ( x )
f (n) ( x ) = lim
∆x →0 ∆x
Remarks
1 The n in f (n) is called the order of the derivative.
Remarks

2 The derivative of a function f is also called the first derivative of f .
Remarks

3 The function f is sometimes written as f (0) ( x ).
Remarks

4 Other notations:
Remarks

4 Other notations: Dx n [ f ( x )] ,
Remarks

dn y
4 Other notations: Dx n [ f ( x )] , ,
dx n
Remarks

dn y dn
4 Other notations: Dx n [ f ( x )] , n
, [ f ( x )] ,
dx dx n
Remarks

dn y dn
4 Other notations: Dx n [ f ( x )] , n
, [ f ( x )] , y(n)
dx dx n
Example
Find f (n) ( x ) for all n ∈ N where f ( x ) = x6 − x4 − 3x3 + 2x2 − 4.
Example
Solution. We differentiate repeatedly and obtain
Example
f 0 (x) = 6x5 − 4x3 − 9x2 + 4x,
Example
f 0 (x) = 6x5 − 4x3 − 9x2 + 4x,

f 00 ( x ) = 30x4 − 12x2 − 18x + 4,
Example
f 0 (x) = 6x5 − 4x3 − 9x2 + 4x,

f 00 ( x ) = 30x4 − 12x2 − 18x + 4,
f 000 ( x ) = 120x3 − 24x − 18,
Example
f 0 (x) = 6x5 − 4x3 − 9x2 + 4x,

f 00 ( x ) = 30x4 − 12x2 − 18x + 4,
f 000 ( x ) = 120x3 − 24x − 18,
f (4) ( x ) = 360x2 − 24,
Example
f 0 (x) = 6x5 − 4x3 − 9x2 + 4x,

f 00 ( x ) = 30x4 − 12x2 − 18x + 4,
f 000 ( x ) = 120x3 − 24x − 18,
f (4) ( x ) = 360x2 − 24,
f (5) ( x ) = 720x,
Example
f 0 (x) = 6x5 − 4x3 − 9x2 + 4x,

f 00 ( x ) = 30x4 − 12x2 − 18x + 4,
f 000 ( x ) = 120x3 − 24x − 18,
f (4) ( x ) = 360x2 − 24,
f (5) ( x ) = 720x,
f (6) ( x ) = 720,
Example
f 0 (x) = 6x5 − 4x3 − 9x2 + 4x,

f 00 ( x ) = 30x4 − 12x2 − 18x + 4,
f 000 ( x ) = 120x3 − 24x − 18,
f (4) ( x ) = 360x2 − 24,
f (5) ( x ) = 720x,
f (6) ( x ) = 720,
(n)
f ( x ) = 0 for all n ≥ 7.
Example
√
Find f (4) ( x ) if f ( x ) = 2x − 3.
Example
√
Find f (4) ( x ) if f ( x ) = 2x − 3.
Solution.
f 0 (x)
Example
√
Find f (4) ( x ) if f ( x ) = 2x − 3.
Solution.
1 1
f 0 (x) = (2x − 3)− 2 · (2)
2
Example
√
Find f (4) ( x ) if f ( x ) = 2x − 3.
Solution.
1 1 1
f 0 (x) = (2x − 3)− 2 · (2) = (2x − 3)− 2
2
Example
√
Find f (4) ( x ) if f ( x ) = 2x − 3.
Solution.
1 1 1
f 0 (x) = (2x − 3)− 2 · (2) = (2x − 3)− 2
2
f 00 ( x )
Example
√
Find f (4) ( x ) if f ( x ) = 2x − 3.
Solution.
1 1 1
f 0 (x) = (2x − 3)− 2 · (2) = (2x − 3)− 2
2
1 3
f 00 ( x ) = − (2x − 3)− 2 · (2)
2
Example
√
Find f (4) ( x ) if f ( x ) = 2x − 3.
Solution.
1 1 1
f 0 (x) = (2x − 3)− 2 · (2) = (2x − 3)− 2
2
1 3 3
f 00 ( x ) = − (2x − 3)− 2 · (2) = − (2x − 3)− 2
2
Example
√
Find f (4) ( x ) if f ( x ) = 2x − 3.
Solution.
1 1 1
f 0 (x) = (2x − 3)− 2 · (2) = (2x − 3)− 2
2
1 3 3
f 00 ( x ) = − (2x − 3)− 2 · (2) = − (2x − 3)− 2
2
f 000 ( x )
Example
√
Find f (4) ( x ) if f ( x ) = 2x − 3.
Solution.
1 1 1
f 0 (x) = (2x − 3)− 2 · (2) = (2x − 3)− 2
2
1 3 3
f 00 ( x ) = − (2x − 3)− 2 · (2) = − (2x − 3)− 2
2
3 5
f 000 ( x ) = (2x − 3)− 2 · (2)
2
Example
√
Find f (4) ( x ) if f ( x ) = 2x − 3.
Solution.
1 1 1
f 0 (x) = (2x − 3)− 2 · (2) = (2x − 3)− 2
2
1 3 3
f 00 ( x ) = − (2x − 3)− 2 · (2) = − (2x − 3)− 2
2
3 5 5
f 000 ( x ) = (2x − 3)− 2 · (2) = 3 (2x − 3)− 2
2
Example
√
Find f (4) ( x ) if f ( x ) = 2x − 3.
Solution.
1 1 1
f 0 (x) = (2x − 3)− 2 · (2) = (2x − 3)− 2
2
1 3 3
f 00 ( x ) = − (2x − 3)− 2 · (2) = − (2x − 3)− 2
2
3 5 5
f 000 ( x ) = (2x − 3)− 2 · (2) = 3 (2x − 3)− 2
2
f (4) ( x )
Example
√
Find f (4) ( x ) if f ( x ) = 2x − 3.
Solution.
1 1 1
f 0 (x) = (2x − 3)− 2 · (2) = (2x − 3)− 2
2
1 3 3
f 00 ( x ) = − (2x − 3)− 2 · (2) = − (2x − 3)− 2
2
3 5 5
f 000 ( x ) = (2x − 3)− 2 · (2) = 3 (2x − 3)− 2
2
15 7
f (4) ( x ) = − (2x − 3)− 2 · (2)
2
Example
√
Find f (4) ( x ) if f ( x ) = 2x − 3.
Solution.
1 1 1
f 0 (x) = (2x − 3)− 2 · (2) = (2x − 3)− 2
2
1 3 3
f 00 ( x ) = − (2x − 3)− 2 · (2) = − (2x − 3)− 2
2
3 5 5
f 000 ( x ) = (2x − 3)− 2 · (2) = 3 (2x − 3)− 2
2
15 7 7
f (4) ( x ) = − (2x − 3)− 2 · (2) = −15 (2x − 3)− 2
2
For today
Equations in two variables are used to define a function explicitly or implicitly.
• y = x2 − 1
• y = x2 − 1 defines f ( x ) = x2 − 1 (explicitly).
• y2 = x + 1
• y2 = x + 1 defines two functions of x: (implicitly)

√
f1 (x) = x + 1

√
f1 (x) = x + 1
√
f 2 ( x ) = − x + 1.
dy
Consider: If x3 − 2xy − 3y − 6 = 0, how do we find ?
dx
dy
dx
Solution
From x3 − 2xy − 3y − 6 = 0, we can define y explicitly in terms of x as
dy
dx
Solution
x3 − 6
y= .
2x + 3
dy
dx
Solution
x3 − 6
y= .
2x + 3
dy
Thus,
dx
dy
dx
Solution
x3 − 6
y= .
2x + 3
dy (2x + 3)(3x2 ) − ( x3 − 6)(2)

Thus, =
dx (2x + 3)2
dy
dx
Solution
x3 − 6
y= .
2x + 3
dy (2x + 3)(3x2 ) − ( x3 − 6)(2) 4x3 + 9x2 + 12

Thus, = = .
dx (2x + 3)2 (2x + 3)2
Question: Suppose we have x4 y3 − 7xy = 7 or tan( x2 − 2xy) = y.

dy
How do we find ?
dx
Question: Suppose we have x4 y3 − 7xy = 7 or tan( x2 − 2xy) = y.

dy
How do we find ?
dx
dy
To find using implicit differentiation, we
dx
dy
dx
1 think of the variable y as a differentiable function of the variable x,
dy
dx
2 differentiate both sides of the equation, using the chain rule where necessary,
and
dy
dx
2 differentiate both sides of the equation, using the chain rule where necessary,
and
dy
3 solve for .
dx
Example
dy
Find if x3 − 2xy − 3y − 6 = 0.
dx
Example
dy
Find if x3 − 2xy − 3y − 6 = 0.
dx
Solution.
Dx [ x3 − 2xy − 3y − 6] = D x [0]
Example
dy
Find if x3 − 2xy − 3y − 6 = 0.
dx
Solution.
Dx [ x3 − 2xy − 3y − 6] = D x [0]
3
Dx [ x ] − Dx [2xy] − Dx [3y] − Dx [6] = D x [0]
Example
dy
Find if x3 − 2xy − 3y − 6 = 0.
dx
Solution.
Dx [ x3 − 2xy − 3y − 6] = D x [0]
3
Dx [ x ] − Dx [2xy] − Dx [3y] − Dx [6] = D x [0]
3x2
Example
dy
Find if x3 − 2xy − 3y − 6 = 0.
dx
Solution.
Dx [ x3 − 2xy − 3y − 6] = D x [0]
3
Dx [ x ] − Dx [2xy] − Dx [3y] − Dx [6] = D x [0]
3x2 − 2
Example
dy
Find if x3 − 2xy − 3y − 6 = 0.
dx
Solution.
Dx [ x3 − 2xy − 3y − 6] = D x [0]
3
Dx [ x ] − Dx [2xy] − Dx [3y] − Dx [6] = D x [0]

3x2 − 2 1
Example
dy
Find if x3 − 2xy − 3y − 6 = 0.
dx
Solution.
Dx [ x3 − 2xy − 3y − 6] = D x [0]
3
Dx [ x ] − Dx [2xy] − Dx [3y] − Dx [6] = D x [0]

3x2 − 2 1 · y
Example
dy
Find if x3 − 2xy − 3y − 6 = 0.
dx
Solution.
Dx [ x3 − 2xy − 3y − 6] = D x [0]
3
Dx [ x ] − Dx [2xy] − Dx [3y] − Dx [6] = D x [0]

3x2 − 2 1 · y + x
Example
dy
Find if x3 − 2xy − 3y − 6 = 0.
dx
Solution.
Dx [ x3 − 2xy − 3y − 6] = D x [0]
3
Dx [ x ] − Dx [2xy] − Dx [3y] − Dx [6] = D x [0]
dy
3x2 − 2 1 · y + x ·
dx
Example
dy
Find if x3 − 2xy − 3y − 6 = 0.
dx
Solution.
Dx [ x3 − 2xy − 3y − 6] = D x [0]
3
Dx [ x ] − Dx [2xy] − Dx [3y] − Dx [6] = D x [0]
dy
3x2 − 2 1 · y + x · −3
dx
Example
dy
Find if x3 − 2xy − 3y − 6 = 0.
dx
Solution.
Dx [ x3 − 2xy − 3y − 6] = D x [0]
3
Dx [ x ] − Dx [2xy] − Dx [3y] − Dx [6] = D x [0]
dy dy
3x2 − 2 1 · y + x · −3·
dx dx
Example
dy
Find if x3 − 2xy − 3y − 6 = 0.
dx
Solution.
Dx [ x3 − 2xy − 3y − 6] = D x [0]
3
Dx [ x ] − Dx [2xy] − Dx [3y] − Dx [6] = D x [0]
dy dy
3x2 − 2 1 · y + x · −3· −0
dx dx
Example
dy
Find if x3 − 2xy − 3y − 6 = 0.
dx
Solution.
Dx [ x3 − 2xy − 3y − 6] = D x [0]
3
Dx [ x ] − Dx [2xy] − Dx [3y] − Dx [6] = D x [0]
dy dy
3x2 − 2 1 · y + x · −3· −0 = 0
dx dx
Example
dy
Find if x3 − 2xy − 3y − 6 = 0.
dx
Solution.
Dx [ x3 − 2xy − 3y − 6] = D x [0]
3
Dx [ x ] − Dx [2xy] − Dx [3y] − Dx [6] = D x [0]
dy dy
3x2 − 2 1 · y + x · −3· −0 = 0
dx dx
dy dy
3x2 − 2y − 2x · −3· = 0
dx dx
Example
dy
Find if x3 − 2xy − 3y − 6 = 0.
dx
Solution.
Dx [ x3 − 2xy − 3y − 6] = D x [0]
3
Dx [ x ] − Dx [2xy] − Dx [3y] − Dx [6] = D x [0]
dy dy
3x2 − 2 1 · y + x · −3· −0 = 0
dx dx
dy dy
3x2 − 2y − 2x · −3· = 0
dx dx
dy dy
2x · +3· = 3x2 − 2y
dx dx
Example
dy
Find if x3 − 2xy − 3y − 6 = 0.
dx
Solution.
Dx [ x3 − 2xy − 3y − 6] = D x [0]
3
Dx [ x ] − Dx [2xy] − Dx [3y] − Dx [6] = D x [0]
dy dy
3x2 − 2 1 · y + x · −3· −0 = 0
dx dx
dy dy
3x2 − 2y − 2x · −3· = 0
dx dx
dy dy
2x · +3· = 3x2 − 2y
dx dx
dy
(2x + 3) = 3x2 − 2y
dx
dy 3x2 − 2y
=
dx 2x + 3
dy 3x2 − 2y
=
dx 2x + 3
x −6 3
3x2 − 2 · 2x +3
=
2x + 3
dy 3x2 − 2y
=
dx 2x + 3
x −6 3
3x2 − 2 · 2x +3
=
2x + 3
3x2 · (2x + 3) − 2( x3 − 6)
=
(2x + 3)2
dy 3x2 − 2y
=
dx 2x + 3
x −6 3
3x2 − 2 · 2x +3
=
2x + 3
3x2 · (2x + 3) − 2( x3 − 6)
=
(2x + 3)2
4x3 + 9x2 + 12
= .
(2x + 3)2
Example
dy
Find if x4 y3 − 7xy = 7.
dx
Example
dy
dx
Solution.
Dx [ x4 y3 − 7xy] = D x [7]
Example
dy
dx
Solution.
Dx [ x4 y3 − 7xy] = D x [7]
4x3 y3
Example
dy
dx
Solution.
Dx [ x4 y3 − 7xy] = D x [7]
4x3 y3 + 3x4 y2
Example
dy
dx
Solution.
Dx [ x4 y3 − 7xy] = D x [7]
dy
4x3 y3 + 3x4 y2 ·
dx
Example
dy
dx
Solution.
Dx [ x4 y3 − 7xy] = D x [7]
dy
4x3 y3 + 3x4 y2 · −7
dx
Example
dy
dx
Solution.
Dx [ x4 y3 − 7xy] = D x [7]
dy
4x3 y3 + 3x4 y2 · −7 y
dx
Example
dy
dx
Solution.
Dx [ x4 y3 − 7xy] = D x [7]
dy
4x3 y3 + 3x4 y2 · −7 y+x
dx
Example
dy
dx
Solution.
Dx [ x4 y3 − 7xy] = D x [7]
dy dy
4x3 y3 + 3x4 y2 · −7 y+x·
dx dx
Example
dy
dx
Solution.
Dx [ x4 y3 − 7xy] = D x [7]
dy dy
4x3 y3 + 3x4 y2 · −7 y+x· = 0
dx dx
Example
dy
dx
Solution.
Dx [ x4 y3 − 7xy] = D x [7]
dy dy
4x3 y3 + 3x4 y2 · −7 y+x· = 0
dx dx
dy −4x3 y3 + 7y
= .
dx 3x4 y2 − 7x
Example
dy
Find if tan( x2 − 2xy) = y.
dx
Example
dy
dx
Solution.
Dx [tan( x2 − 2xy)] = Dx [y]
Example
dy
dx
Solution.
Dx [tan( x2 − 2xy)] = Dx [y]
sec2 ( x2 − 2xy)
Example
dy
dx
Solution.
Dx [tan( x2 − 2xy)] = Dx [y]

h
sec2 ( x2 − 2xy) · 2x
Example
dy
dx
Solution.
Dx [tan( x2 − 2xy)] = Dx [y]

h
sec2 ( x2 − 2xy) · 2x − 2
Example
dy
dx
Solution.
Dx [tan( x2 − 2xy)] = Dx [y]

h
sec2 ( x2 − 2xy) · 2x − 2 y
Example
dy
dx
Solution.
Dx [tan( x2 − 2xy)] = Dx [y]

h
sec2 ( x2 − 2xy) · 2x − 2 y + x
Example
dy
dx
Solution.
Dx [tan( x2 − 2xy)] = Dx [y]

h dy i
sec2 ( x2 − 2xy) · 2x − 2 y + x ·
dx
Example
dy
dx
Solution.
Dx [tan( x2 − 2xy)] = Dx [y]

h dy i dy
sec2 ( x2 − 2xy) · 2x − 2 y + x · =
dx dx
Example
dy
dx
Solution.
Dx [tan( x2 − 2xy)] = Dx [y]

h dy i dy
sec2 ( x2 − 2xy) · 2x − 2 y + x · =
dx dx
Thus,
dy 2x sec2 ( x2 − 2xy) − 2y sec2 ( x2 − 2xy)

= .
dx 1 + 2x sec2 ( x2 − 2xy)
Example
d2 y
Determine if xy2 = y − 2.
dx2
Example
d2 y
dx2
Solution.
Dx [ xy2 ] = D x [ y − 2]
Example
d2 y
dx2
Solution.
Dx [ xy2 ] = D x [ y − 2]
y2
Example
d2 y
dx2
Solution.
Dx [ xy2 ] = D x [ y − 2]
y2 + x
Example
d2 y
dx2
Solution.
Dx [ xy2 ] = D x [ y − 2]
y2 + x · 2y
Example
d2 y
dx2
Solution.
Dx [ xy2 ] = D x [ y − 2]
dy
y2 + x · 2y
dx
Example
d2 y
dx2
Solution.
Dx [ xy2 ] = D x [ y − 2]
dy dy
y2 + x · 2y =
dx dx
Example
d2 y
dx2
Solution.
Dx [ xy2 ] = D x [ y − 2]
dy dy
y2 + x · 2y =
dx dx
dy y2
=
dx 1 − 2xy
dy y2
We have = .
dx 1 − 2xy
dy y2
We have = .
dx 1 − 2xy
d2 y
=⇒
dx2
dy y2
We have = .
dx 1 − 2xy
d2 y
=⇒ =
dx2
dy y2
We have = .
dx 1 − 2xy
d2 y (1 − 2xy)
=⇒ =
dx2
dy y2
We have = .
dx 1 − 2xy
dy
d2 y (1 − 2xy)(2y dx )
=⇒ =
dx2
dy y2
We have = .
dx 1 − 2xy
dy
d2 y (1 − 2xy)(2y dx ) − ( y2 )
=⇒ =
dx2
dy y2
We have = .
dx 1 − 2xy
dy dy
d2 y (1 − 2xy)(2y dx ) − (y2 )[−2(y + x dx )]
=⇒ =
dx2
dy y2
We have = .
dx 1 − 2xy
dy dy
d2 y (1 − 2xy)(2y dx ) − (y2 )[−2(y + x dx )]
=⇒ =
dx2 (1 − 2xy)2
dy y2
We have = .
dx 1 − 2xy
dy dy
d2 y (1 − 2xy)(2y dx ) − (y2 )[−2(y + x dx )]
=⇒ =
dx2 (1 − 2xy)2
2 y2
h i h i
(1 − 2xy) 2y · 1−y2xy − (y2 ) −2 y + x · 1−2xy
= .
(1 − 2xy)2
For today
√
3
Suppose we are asked to find 8.03.
√
3
Suppose we are asked to find 8.03. Most (if not all) of us cannot give an
accurate decimal expansion of this.
√
3
Suppose we are asked to find 8.03. Most (if not all) of us cannot give an
accurate decimal expansion of this. We will give an estimate using the fact that a
tangent line to a curve at a point P approximates the curve near P. That is, R
approximates Q when x is near x0 .
Q
R
f (x)
P
f ( x0 )
S
| {z }
dx =∆x
x0 x = x0 + dx
y = f (x)
Recall:
f ( x ) − f ( x0 )
f 0 ( x0 ) = lim
x → x0 x − x0
Recall:
f ( x ) − f ( x0 )
f 0 ( x0 ) = lim
x → x0 x − x0
∆y
= lim ,
∆x →0 ∆x
where ∆x = f ( x ) − f ( x0 ) and ∆x = x − x0 .
Recall:
f ( x ) − f ( x0 )
f 0 ( x0 ) = lim
x → x0 x − x0
∆y
= lim ,
∆x →0 ∆x
If ∆x is small enough,
Recall:
f ( x ) − f ( x0 )
f 0 ( x0 ) = lim
x → x0 x − x0
∆y
= lim ,
∆x →0 ∆x
If ∆x is small enough, then
∆y
≈ f 0 ( x0 )
∆x
Recall:
f ( x ) − f ( x0 )
f 0 ( x0 ) = lim
x → x0 x − x0
∆y
= lim ,
∆x →0 ∆x
∆y
≈ f 0 ( x0 ) =⇒ ∆y ≈ f 0 ( x0 )∆x
∆x
Recall:
f ( x ) − f ( x0 )
f 0 ( x0 ) = lim
x → x0 x − x0
∆y
= lim ,
∆x →0 ∆x
∆y
≈ f 0 ( x0 ) =⇒ ∆y ≈ f 0 ( x0 )∆x
∆x
The expression on the right, which approximates the change in y due to a change
in x, is called a differential.
Definitions
Let the function y = f ( x ) be differentiable at x.
Definitions
1 The differential dx of the independent variable x denotes an arbitrary
increment of x.
Definitions
1 The differential dx of the independent variable x denotes an arbitrary
increment of x.
2 The differential dy of the dependent variable y associated with x is given

by dy = f 0 ( x )dx.
Example
Find dy if
y = x5 − x3 + 2x.
Example
Find dy if
y = x5 − x3 + 2x.
Solution.
dy
Example
Find dy if
y = x5 − x3 + 2x.
Solution.
dy = (5x4 − 3x2 + 2)
Example
Find dy if
y = x5 − x3 + 2x.
Solution.
dy = (5x4 − 3x2 + 2) dx
Example
Find dy if
√
y= x3 + 3x2 .
Example
Find dy if
√
y= x3 + 3x2 .
Solution.
dy
Example
Find dy if
√
y= x3 + 3x2 .
Solution.
1
dy = √
2 x + 3x2
3
Example
Find dy if
√
y= x3 + 3x2 .
Solution.
1
dy = √ · (3x2 + 6x )
2 x + 3x2
3
Example
Find dy if
√
y= x3 + 3x2 .
Solution.
1
dy = √ · (3x2 + 6x ) dx
2 x + 3x2
3
Let dx = ∆x = x − x0 , ∆y = f ( x ) − f ( x0 ).
If dx ≈ 0
Let dx = ∆x = x − x0 , ∆y = f ( x ) − f ( x0 ).
If dx ≈ 0
=⇒ ∆y ≈ f 0 ( x0 )dx
Let dx = ∆x = x − x0 , ∆y = f ( x ) − f ( x0 ).
If dx ≈ 0
=⇒ ∆y ≈ f 0 ( x0 )dx = dy
=⇒ f ( x ) ≈ f ( x0 ) + f 0 ( x0 )( x − x0 )
Let dx = ∆x = x − x0 , ∆y = f ( x ) − f ( x0 ).
If dx ≈ 0
=⇒ ∆y ≈ f 0 ( x0 )dx = dy
=⇒ f ( x ) ≈ f ( x0 ) + f 0 ( x0 )( x − x0 )
Equation of the tangent line to the graph of y = f ( x ) at the point ( x0 , f ( x0 )):
Let dx = ∆x = x − x0 , ∆y = f ( x ) − f ( x0 ).
If dx ≈ 0
=⇒ ∆y ≈ f 0 ( x0 )dx = dy
=⇒ f ( x ) ≈ f ( x0 ) + f 0 ( x0 )( x − x0 )
y − f ( x0 ) = f 0 ( x0 )( x − x0 )
Let dx = ∆x = x − x0 , ∆y = f ( x ) − f ( x0 ).
If dx ≈ 0
=⇒ ∆y ≈ f 0 ( x0 )dx = dy
=⇒ f ( x ) ≈ f ( x0 ) + f 0 ( x0 )( x − x0 )
y − f ( x0 ) = f 0 ( x0 )( x − x0 )
∴
Let dx = ∆x = x − x0 , ∆y = f ( x ) − f ( x0 ).
If dx ≈ 0
=⇒ ∆y ≈ f 0 ( x0 )dx = dy
=⇒ f ( x ) ≈ f ( x0 ) + f 0 ( x0 )( x − x0 )
y − f ( x0 ) = f 0 ( x0 )( x − x0 )
∴ y = f ( x0 ) + f 0 ( x0 )( x − x0 )
Let dx = ∆x = x − x0 , ∆y = f ( x ) − f ( x0 ).
If dx ≈ 0
=⇒ ∆y ≈ f 0 ( x0 )dx = dy
=⇒ f ( x ) ≈ f ( x0 ) + f 0 ( x0 )( x − x0 )
y − f ( x0 ) = f 0 ( x0 )( x − x0 )
∴ y = f ( x0 ) + f 0 ( x0 )( x − x0 )
Thus, when dx ≈ 0, or x ≈ x0 , f ( x ) is approximated by the tangent line to f at x0 .
y = f (x)
`
f (x)
f ( x0 ) + f 0 ( x0 )( x − x0 )
x0 x
Remarks
Remarks
The tangent line L( x ) = f ( x0 ) + f 0 ( x0 )( x − x0 ) is also called the local

linear approximation of f ( x ) at x0 .
Remarks

The local linear approximation is the “best” linear approximation of f near x0 .
Remarks

If dx = ∆x = x − x0 ,
Remarks

If dx = ∆x = x − x0 , then x = x0 + dx.
Remarks

Since f ( x ) ≈ f ( x0 ) + f 0 ( x0 )( x − x0 ), we have
Remarks

Since f ( x ) ≈ f ( x0 ) + f 0 ( x0 )( x − x0 ), we have
f ( x0 + dx ) ≈ f ( x0 ) + f 0 ( x0 )dx
Remarks
dx = ∆x ≈ 0
Remarks
dx = ∆x ≈ 0 =⇒
Remarks
dx = ∆x ≈ 0 =⇒ ∆y ≈ dy
Remarks
dx = ∆x ≈ 0 =⇒ ∆y ≈ dy
dy is easier to compute than ∆y
Remarks
dx = ∆x ≈ 0 =⇒ ∆y ≈ dy
∴ dy is used to approximate ∆y when dx ≈ 0
Remarks
dx = ∆x ≈ 0 =⇒ ∆y ≈ dy

If dx 6= 0,
Remarks
dx = ∆x ≈ 0 =⇒ ∆y ≈ dy

If dx 6= 0, then
dy = f 0 ( x )dx
Remarks
dx = ∆x ≈ 0 =⇒ ∆y ≈ dy

If dx 6= 0, then
dy
dy = f 0 ( x )dx =⇒ = f 0 ( x ).
dx
Remarks
dx = ∆x ≈ 0 =⇒ ∆y ≈ dy

If dx 6= 0, then
dy
dy = f 0 ( x )dx =⇒ = f 0 ( x ).
dx
dy
The symbol may be interpreted as:
dx
Remarks
dx = ∆x ≈ 0 =⇒ ∆y ≈ dy

If dx 6= 0, then
dy
dy = f 0 ( x )dx =⇒ = f 0 ( x ).
dx
dy
dx
the derivative of y = f ( x ) with respect to x
Remarks
dx = ∆x ≈ 0 =⇒ ∆y ≈ dy

If dx 6= 0, then
dy
dy = f 0 ( x )dx =⇒ = f 0 ( x ).
dx
dy
dx
the derivative of y = f ( x ) with respect to x
the quotient of the differential of y by the differential of x
Example
√
3
Find the local linear approximation of f ( x ) = x at x0 = 8 and use this to
√
3
estimate 8.03.
Example
√
3
√
3
estimate 8.03.
Solution.
1
We have f 0 ( x ) = √
3
.
3 x2
Example
√
3
√
3
estimate 8.03.
Solution.
1
We have f 0 ( x ) = √
3
.
3 x2
L( x ) = f ( x0 ) + f 0 ( x0 )( x − x0 )
Example
√
3
√
3
estimate 8.03.
Solution.
1
We have f 0 ( x ) = √
3
.
3 x2
L( x ) = f ( x0 ) + f 0 ( x0 )( x − x0 )
∴ At x0 = 8:
Example
√
3
√
3
estimate 8.03.
Solution.
1
We have f 0 ( x ) = √
3
.
3 x2
L( x ) = f ( x0 ) + f 0 ( x0 )( x − x0 )
∴ At x0 = 8:
L( x )
Example
√
3
√
3
estimate 8.03.
Solution.
1
We have f 0 ( x ) = √
3
.
3 x2
L( x ) = f ( x0 ) + f 0 ( x0 )( x − x0 )
∴ At x0 = 8:
L ( x ) = f (8)
Example
√
3
√
3
estimate 8.03.
Solution.
1
We have f 0 ( x ) = √
3
.
3 x2
L( x ) = f ( x0 ) + f 0 ( x0 )( x − x0 )
∴ At x0 = 8:
L ( x ) = f (8) + f 0 (8)
Example
√
3
√
3
estimate 8.03.
Solution.
1
We have f 0 ( x ) = √
3
.
3 x2
L( x ) = f ( x0 ) + f 0 ( x0 )( x − x0 )
∴ At x0 = 8:
L( x ) = f (8) + f 0 (8)( x − 8)
Example
√
3
√
3
estimate 8.03.
Solution.
1
We have f 0 ( x ) = √
3
.
3 x2
L( x ) = f ( x0 ) + f 0 ( x0 )( x − x0 )
∴ At x0 = 8:
L( x ) = f (8) + f 0 (8)( x − 8) = 2
Example
√
3
√
3
estimate 8.03.
Solution.
1
We have f 0 ( x ) = √
3
.
3 x2
L( x ) = f ( x0 ) + f 0 ( x0 )( x − x0 )
∴ At x0 = 8:
1
L( x ) = f (8) + f 0 (8)( x − 8) = 2 +
12
Example
√
3
√
3
estimate 8.03.
Solution.
1
We have f 0 ( x ) = √
3
.
3 x2
L( x ) = f ( x0 ) + f 0 ( x0 )( x − x0 )
∴ At x0 = 8:
1
L( x ) = f (8) + f 0 (8)( x − 8) = 2 + ( x − 8).
12
Example
√
3
√
3
estimate 8.03.
Solution.
1
We have f 0 ( x ) = √
3
.
3 x2
L( x ) = f ( x0 ) + f 0 ( x0 )( x − x0 )
∴ At x0 = 8:
1
L( x ) = f (8) + f 0 (8)( x − 8) = 2 + ( x − 8).
12
√
3
Thus, 8.03
Example
√
3
√
3
estimate 8.03.
Solution.
1
We have f 0 ( x ) = √
3
.
3 x2
L( x ) = f ( x0 ) + f 0 ( x0 )( x − x0 )
∴ At x0 = 8:
1
L( x ) = f (8) + f 0 (8)( x − 8) = 2 + ( x − 8).
12
√
3
Thus, 8.03 = f (8.03)
Example
√
3
√
3
estimate 8.03.
Solution.
1
We have f 0 ( x ) = √
3
.
3 x2
L( x ) = f ( x0 ) + f 0 ( x0 )( x − x0 )
∴ At x0 = 8:
1
L( x ) = f (8) + f 0 (8)( x − 8) = 2 + ( x − 8).
12
√
3 1
Thus, 8.03 = f (8.03) ≈ L(8.03) = 2 + (8.03 − 8)
12
Example
√
3
√
3
estimate 8.03.
Solution.
1
We have f 0 ( x ) = √
3
.
3 x2
L( x ) = f ( x0 ) + f 0 ( x0 )( x − x0 )
∴ At x0 = 8:
1
L( x ) = f (8) + f 0 (8)( x − 8) = 2 + ( x − 8).
12
√
3 1
Thus, 8.03 = f (8.03) ≈ L(8.03) = 2 + (8.03 − 8) = 2.0025.
12
Example
√
3
Approximate 27.027.
Example
√
3
Approximate 27.027.
Solution.
Example
√
3
Approximate 27.027.
Solution.
√
3
Let f ( x ) = x.
Example
√
3
Approximate 27.027.
Solution.
√ 1
Let f ( x ) = 3
x. Then f 0 ( x ) = √
3 2
.
3 x
Example
√
3
Approximate 27.027.
Solution.
√ 1
Let f ( x ) = 3
x. Then f 0 ( x ) = √
3 2
.
3 x
f ( x0 + dx ) ≈ f ( x0 ) + f 0 ( x0 )dx
Example
√
3
Approximate 27.027.
Solution.
√ 1
Let f ( x ) = 3
x. Then f 0 ( x ) = √
3 2
.
3 x
f ( x0 + dx ) ≈ f ( x0 ) + f 0 ( x0 )dx
Thus,
√
3
27.027
Example
√
3
Approximate 27.027.
Solution.
√ 1
Let f ( x ) = 3
x. Then f 0 ( x ) = √
3 2
.
3 x
f ( x0 + dx ) ≈ f ( x0 ) + f 0 ( x0 )dx
Thus,
√
3
27.027 = f (27 + 0.027)
Example
√
3
Approximate 27.027.
Solution.
√ 1
Let f ( x ) = 3
x. Then f 0 ( x ) = √
3 2
.
3 x
f ( x0 + dx ) ≈ f ( x0 ) + f 0 ( x0 )dx
Thus,
√
3
27.027 = f (27 + 0.027)
≈ f (27) + f 0 (27) · (0.027)
Example
√
3
Approximate 27.027.
Solution.
√ 1
Let f ( x ) = 3
x. Then f 0 ( x ) = √
3 2
.
3 x
f ( x0 + dx ) ≈ f ( x0 ) + f 0 ( x0 )dx
Thus,
√
3
27.027 = f (27 + 0.027)
≈ f (27) + f 0 (27) · (0.027)
= 3
Example
√
3
Approximate 27.027.
Solution.
√ 1
Let f ( x ) = 3
x. Then f 0 ( x ) = √
3 2
.
3 x
f ( x0 + dx ) ≈ f ( x0 ) + f 0 ( x0 )dx
Thus,
√
3
27.027 = f (27 + 0.027)
≈ f (27) + f 0 (27) · (0.027)
1
= 3+
3·9
Example
√
3
Approximate 27.027.
Solution.
√ 1
Let f ( x ) = 3
x. Then f 0 ( x ) = √
3 2
.
3 x
f ( x0 + dx ) ≈ f ( x0 ) + f 0 ( x0 )dx
Thus,
√
3
27.027 = f (27 + 0.027)
≈ f (27) + f 0 (27) · (0.027)
1
= 3+ · (0.027)
3·9
Example
√
3
Approximate 27.027.
Solution.
√ 1
Let f ( x ) = 3
x. Then f 0 ( x ) = √
3 2
.
3 x
f ( x0 + dx ) ≈ f ( x0 ) + f 0 ( x0 )dx
Thus,
√
3
27.027 = f (27 + 0.027)
≈ f (27) + f 0 (27) · (0.027)
1
= 3+ · (0.027)
3·9
= 3.01.
Example
√
Approximate 15.96.
Example
√
Approximate 15.96.
Solution.
Example
√
Approximate 15.96.
Solution.
√
Let f ( x ) = x.
Example
√
Approximate 15.96.
Solution.
√ 1
Let f ( x ) = x. Then f 0 ( x ) = √ .
2 x
Example
√
Approximate 15.96.
Solution.
√ 1
Let f ( x ) = x. Then f 0 ( x ) = √ .
2 x
Thus,
√
15.96
Example
√
Approximate 15.96.
Solution.
√ 1
Let f ( x ) = x. Then f 0 ( x ) = √ .
2 x
Thus,
√
15.96 = f (16 − 0.04)
Example
√
Approximate 15.96.
Solution.
√ 1
Let f ( x ) = x. Then f 0 ( x ) = √ .
2 x
Thus,
√
15.96 = f (16 − 0.04)
≈ f (16) + f 0 (16) · (−0.04)
Example
√
Approximate 15.96.
Solution.
√ 1
Let f ( x ) = x. Then f 0 ( x ) = √ .
2 x
Thus,
√
15.96 = f (16 − 0.04)
≈ f (16) + f 0 (16) · (−0.04)
= 4
Example
√
Approximate 15.96.
Solution.
√ 1
Let f ( x ) = x. Then f 0 ( x ) = √ .
2 x
Thus,
√
15.96 = f (16 − 0.04)
≈ f (16) + f 0 (16) · (−0.04)
1
= 4+
2·4
Example
√
Approximate 15.96.
Solution.
√ 1
Let f ( x ) = x. Then f 0 ( x ) = √ .
2 x
Thus,
√
15.96 = f (16 − 0.04)
≈ f (16) + f 0 (16) · (−0.04)
1
= 4+ · (−0.04)
2·4
Example
√
Approximate 15.96.
Solution.
√ 1
Let f ( x ) = x. Then f 0 ( x ) = √ .
2 x
Thus,
√
15.96 = f (16 − 0.04)
≈ f (16) + f 0 (16) · (−0.04)
1
= 4+ · (−0.04)
2·4
= 4 − 0.005
Example
√
Approximate 15.96.
Solution.
√ 1
Let f ( x ) = x. Then f 0 ( x ) = √ .
2 x
Thus,
√
15.96 = f (16 − 0.04)
≈ f (16) + f 0 (16) · (−0.04)
1
= 4+ · (−0.04)
2·4
= 4 − 0.005
= 3.995.
Example
1
A ball 10 inches in diameter is to be covered by a rubber material which is 16 in
thick. Use differentials to estimate the volume of the rubber material that will be
used.
Example
1
used.
Solution.
Volume of a sphere with radius r:
Example
1
used.
Solution.
4 3
Volume of a sphere with radius r: V (r ) = πr
3
Example
1
used.
Solution.
4 3
Volume of a sphere with radius r: V (r ) = πr =⇒ dV
3
Example
1
used.
Solution.
4 3
Volume of a sphere with radius r: V (r ) = πr =⇒ dV = 4πr2 dr
3
Example
1
used.
Solution.
4 3
3
Volume of rubber material
Example
1
used.
Solution.
4 3
3
1
Volume of rubber material = V (5 + 16 ) − V (5)
Example
1
used.
Solution.
4 3
3
1
= ∆V
Example
1
used.
Solution.
4 3
3
1
= ∆V r = 5, dr = 1
16
Example
1
used.
Solution.
4 3
3
1
= ∆V r = 5, dr = 1
16
≈ dV
Example
1
used.
Solution.
4 3
3
1
= ∆V r = 5, dr = 1
16
≈ dV
= 4πr2 dr
Example
1
used.
Solution.
4 3
3
1
= ∆V r = 5, dr = 1
16
≈ dV
= 4πr2 dr
1
= 4π (5)2 · ( 16 )
Example
1
used.
Solution.
4 3
3
1
= ∆V r = 5, dr = 1
16
≈ dV
= 4πr2 dr
1
= 4π (5)2 · ( 16 )
25π
= 4 in3
Example
A metal rod 15 cm long and 8 cm in diameter is to be insulated, except for the
ends, with a material 0.001 cm thick. Use differentials to estimate the volume of
the insulation.
Example
the insulation.
Solution.
Volume of the rod:
Example
the insulation.
Solution.
Volume of the rod: V = πr2 h
Example
the insulation.
Solution.
Volume of the rod: V = πr2 h = 15πr2
Example
the insulation.
Solution.
Volume of the rod: V = πr2 h = 15πr2 =⇒ dV
Example
the insulation.
Solution.
Volume of the rod: V = πr2 h = 15πr2 =⇒ dV = 30πr dr
Example
the insulation.
Solution.
Volume of the insulation
Example
the insulation.
Solution.
Volume of the insulation = V (4 + 0.001) − V (4)
Example
the insulation.
Solution.

= ∆V
Example
the insulation.
Solution.

= ∆V r = 4, dr = 0.001
Example
the insulation.
Solution.

= ∆V r = 4, dr = 0.001
≈ dV
Example
the insulation.
Solution.

= ∆V r = 4, dr = 0.001
≈ dV
= 30π (4) · (0.001)
Example
the insulation.
Solution.

= ∆V r = 4, dr = 0.001
≈ dV
= 30π (4) · (0.001)
= 0.12π cm3 .
Example
Suppose that the side of a square is measured with a ruler to be 8 inches with a
1
measurement error of at most ± 64 of an inch. Estimate the error in the computed
area of the square.
Example
1
area of the square.
Solution.
Area of square with side x:
Example
1
area of the square.
Solution.
Area of square with side x: A( x ) = x2
Example
1
area of the square.
Solution.
Area of square with side x: A( x ) = x2 =⇒ dA
Example
1
area of the square.
Solution.
Area of square with side x: A( x ) = x2 =⇒ dA = 2x dx
Example
1
area of the square.
Solution.
1
Measurement error of at most ± 64
Example
1
area of the square.
Solution.
1 1
Measurement error of at most ± 64 =⇒ |dx | = 64
Example
1
area of the square.
Solution.
1 1
Error in the computed area
Example
1
area of the square.
Solution.
1 1
Error in the computed area = ∆A
Example
1
area of the square.
Solution.
1 1
We are computing error from A(8), so we take x = 8.
Example
1
area of the square.
Solution.
1 1
|∆A|
Example
1
area of the square.
Solution.
1 1
|∆A| ≈ |dA|
Example
1
area of the square.
Solution.
1 1
|∆A| ≈ |dA| = |2x | · |dx |
Example
1
area of the square.
Solution.
1 1
|∆A| ≈ |dA| = |2x | · |dx | = 2(8)
Example
1
area of the square.
Solution.
1 1
|∆A| ≈ |dA| = |2x | · |dx | = 2(8)( 64

1
)
Example
1
area of the square.
Solution.
1 1
|∆A| ≈ |dA| = |2x | · |dx | = 2(8)( 64

1
)= 1
4
Example
1
area of the square.
Solution.
1 1
|∆A| ≈ |dA| = |2x | · |dx | = 2(8)( 64

1
)= 1
4
∴ The propagated error in the computed area is at most ± 41 of a square inch.
Exercises
1 Determine Dx4 [ cos(4x ) ].
dy
2 Find if
dx
sin( x + y) = xy2 − 2x3 .
√
4
3 Approximate 16.08.

M53 Lec2.3 Higher Order Derivatives Implicit Differentiation Linear Approximation

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Higher Order Derivatives

Institute of Mathematics (UP Diliman)

1 Higher Order Derivatives

3 Local Linear Approximation and Differentials

1 Higher Order Derivatives

3 Local Linear Approximation and Differentials

If f 0 is differentiable, then the derivative of f 0 is called the second derivative of f

If f 0 is differentiable, then the derivative of f 0 is called the second derivative of f

We can continue to obtain the third derivative, f 000 ,

If f 0 is differentiable, then the derivative of f 0 is called the second derivative of f

If f 0 is differentiable, then the derivative of f 0 is called the second derivative of f

If f 0 is differentiable, then the derivative of f 0 is called the second derivative of f

If f 0 is differentiable, then the derivative of f 0 is called the second derivative of f

1 The n in f (n) is called the order of the derivative.

1 The n in f (n) is called the order of the derivative.

1 The n in f (n) is called the order of the derivative.

1 The n in f (n) is called the order of the derivative.

1 The n in f (n) is called the order of the derivative.

1 The n in f (n) is called the order of the derivative.

1 The n in f (n) is called the order of the derivative.

1 The n in f (n) is called the order of the derivative.

Solution. We differentiate repeatedly and obtain

Solution. We differentiate repeatedly and obtain

f 0 (x) = 6x5 − 4x3 − 9x2 + 4x,

Solution. We differentiate repeatedly and obtain

f 0 (x) = 6x5 − 4x3 − 9x2 + 4x,

Solution. We differentiate repeatedly and obtain

f 0 (x) = 6x5 − 4x3 − 9x2 + 4x,

Solution. We differentiate repeatedly and obtain

f 0 (x) = 6x5 − 4x3 − 9x2 + 4x,

Solution. We differentiate repeatedly and obtain

f 0 (x) = 6x5 − 4x3 − 9x2 + 4x,

Solution. We differentiate repeatedly and obtain

f 0 (x) = 6x5 − 4x3 − 9x2 + 4x,

Solution. We differentiate repeatedly and obtain

f 0 (x) = 6x5 − 4x3 − 9x2 + 4x,

1 Higher Order Derivatives

3 Local Linear Approximation and Differentials

• y2 = x + 1 defines two functions of x: (implicitly)

• y2 = x + 1 defines two functions of x: (implicitly)

• y2 = x + 1 defines two functions of x: (implicitly)

dy (2x + 3)(3x2 ) − ( x3 − 6)(2)

dy (2x + 3)(3x2 ) − ( x3 − 6)(2) 4x3 + 9x2 + 12

Question: Suppose we have x4 y3 − 7xy = 7 or tan( x2 − 2xy) = y.

Question: Suppose we have x4 y3 − 7xy = 7 or tan( x2 − 2xy) = y.

Dx [tan( x2 − 2xy)] = Dx [y]

Dx [tan( x2 − 2xy)] = Dx [y]

Dx [tan( x2 − 2xy)] = Dx [y]

Dx [tan( x2 − 2xy)] = Dx [y]

Dx [tan( x2 − 2xy)] = Dx [y]

Dx [tan( x2 − 2xy)] = Dx [y]

Dx [tan( x2 − 2xy)] = Dx [y]

Dx [tan( x2 − 2xy)] = Dx [y]

Dx [tan( x2 − 2xy)] = Dx [y]

dy 2x sec2 ( x2 − 2xy) − 2y sec2 ( x2 − 2xy)

1 Higher Order Derivatives

3 Local Linear Approximation and Differentials

If ∆x is small enough, then

If ∆x is small enough, then