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# Derivatives

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University of Ottawa

Mathematics

MAT1320

Kirill Zaynullin

Fall

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Calculus and Vectors – How to get an A+
2.1 Derivative Function
A Derivative Function B Differentiability
Given a function y = f (x), the derivative A function y = f (x)is called differentiable atxif f '(xexists.
function of f is a new function called f (f A function y = f (x)is differentiable over an open interval(a,b)if
prime), defined at x by: the function is differentiable at every number in that interval.
Note: The domain of derivative function f is a subset of the
f x ) = limf x+ )h ( f x
h→0 h domain of the original function f : D f ' D f. So a function is
defined over D fbut is differentiable overD f '
C Interpretations of Derivative Function D Notations and Reading
1. The slope of the tangent line to the graph y'= f '(x) [Lagrange Notation]
of y = f (x)at the pointP(a, f (a)is given by
Read: “y prime” or “f prime at x”
m = f '(a).
dy d
2. The instantaneous rate of change in the = f (x) [Leibnitz Notation]
variable y with respect to the variable x , dx dx
Read: “dee y by dee x”
where y = f (x, at x= a is given by:
IRC = f '(a.
dy
f a ) =
dx x=a
Read: “f prime at a, dee y by dee x at x equals a”
E First Principles Ex 1. Use first principles to differentiate each function.
Differentiation is the process to find the a) f (x) = 2x − x
derivative function for a given function.
f (x + h) = 2(x + h)−(x + h)
First Principles is the process of f x+ h) − f x) 2(x +h ) (x + h)3− [2x− x3]
differentiation by computing the limit: f x )= lim = lim
h→0 h h→ 0 h
3 3
f x+ )h ( f x = lim 2(x+ h) 2 x − lim (x+ h) − x
f x ) = lim h→0 h h→0 h
h→0 h
h[( + h)2+ ( + h)x + x2] 2
= lim2 − lim = − x
h→0 h→0 h
∴ f '(x) = 2−3x2
− 3 c) f (x) = ax + b
b) f x = 2
x
−3
f (x h) = 2 f'x )= lim f x + h)− f x) = lim a x + h)+ b − ax+ b
(x + ) h→0 h h→0 h
−3 − 3
− a x + h) +b − ax+ b a x + h)+ b + ax + b
f x +h )− f x) (x + h)2 x2 = lim
f x )= lim = lim h→0 h a x + h)+ b + ax + b
h→0 h h→0 h
1 1 a x + h)+ b − (ax+ b)
2 − 2 2 2 = h→0
(x+ h) x x − (x+ h) h( a x+ h) +b + ax+ b )
= − 3h→0 = −3lh→0 2 2
h h x+ h) x = lim ah = lim a
2 2 2 h→0 h( a x+ h) +b + ax+ b ) h→0 a x+ h) +b + ax +b
= − 3lim x − [x +2 xh+ h ] = −3lim − 2x − h
h→0 h x + h)2x2 h→ 0(x+ h)2 x2 a a
= =
− x 6 a x +0) + b+ ax +b 2 ax + b
= − 3 2 2 = 3
x x x ∴ f 'x = a
6
∴ f'( )= 3 2 ax + b
x
2.1 Derivative Function
©2010 Iulia & Teodoru Gugoiu - Page 1 of 2 Calculus and Vectors – How to get an A+
G Non-Differentiability Notes:
A function is not differentiable at x = a if 1. If a function f is not continuous at x = athen the function f is
f '(adoes not exist. not differentiable at x = a.
2. If a function f is continuous at x = a then the function f may
be or not differentiable at x = a.
H Differentiability Point
If the function y = f (x)is differentiable at
x = a then the tangent line at P(a, f (a))is
unique and not vertical (the slope of the
tangent line is not ∞ or −∞ ).
I Corner Point Ex 2. Find the numbers x where the function y = f (x)(see the
P(a, f (a))is a corner point if there are two graph below) is not differentiable and explain why.
distinct tangent lines at P , one for the left-
hand branch and one for the right-hand
branch. For example:
⎧f ( ), x < a
f( ) = ⎨ 1 and f1'(a) ≠ f2'(a)
⎩f2(x), x > a
J Infinite Slope Point
P(a, f (a)is a infinite slope point if the
tangent line at P is vertical and the function
is increasing or decreasing in the
neighborhood at the of the point P .
f '(a) = ∞or f '(a) = −∞
The function y = f (x) is not differentiable at:
x = −6 (infinite break)
x = −2 (P(−2,4) is a corner point)
x = 0 (jump discontinuity)
x = 2 (P(2,4) is a cusp point)
K Cusp Point x = 4 (removable discontinuity)
P(a, f (a))is a cusp point if the tangent line
x = 6 (P(6,4) is an infinite slope point)
at P is vertical and the function is increasing
on one side of the point P and decreasing on
the other side.
f '(a) = DNE
Reading: Nelson Textbook, Pages 65-72
Homework: Nelson Textbook: Page 73 #1, 6, 7b, 9, 14, 16, 19
2.1 Derivative Function
©2010 Iulia & Teodoru Gugoiu - Page 2 of 2 Calculus and Vectors – How to get an A+
2.2 Derivative of Polynomial Functions
A Power Rule Ex 1. For each case, differentiate.
Consider the power function: y = x , x,n∈R . a) (1)'= (x )'= 0x0−1= 0
Then: b) (x)'= (x )'=1(x1−1) =1(x ) =1
c) (x )'= 5(x5−1) = 5x4
(x )'= nx n−1
'
d n n−1 d) ⎛1 ⎞ =(x −1)' ( 1) −1 1= −x −2= − 1
x = nx ⎜ x⎟ 2
dx ⎝ ⎠ x
'
Some useful specific cases: e) ⎜ 1 ⎟ = (x−7)'= (−7)x −7 1= −7 x−8 = − 7
⎝ x7 ⎠ x 8
(1)' 0 1 1−1 −1
f)( x)' =x )' = 1 x 2 = 1 x 2 = 1 1 = 1
( )' 1 2 2 2 1 2 x
x 2
( x )' 1
1 1 1−1 1 −2 1 1
2 x g) ( x)'= (x )'= x3 = x 3 = =
3 3 2 3 2
3 x3 3 x
3 3 −1
4 3 3 −1 3 3 3
h) ( x )'= (x )'= x 4 = x 4 = =
4 4 1 4 4x
4x 4
π π −1
i)(x )'=πx
B Constant Function Rule Ex 2. Find each derivative function:
Let consider a constant function: f (x) = c, c∈R . a) (−2)'= 0
Then: b) ( π )'= 0
(c)'= 0
d
dx c = 0
C Constant Multiple Rule Ex 3. Differentiate each expression:
Let consider g(x) = cf (x. Then: a) − 2x 3
(−2x )'= (−2)(x )'= (−2)(3)x 3−1 = −6x 2
[cf( )]'= cf'( )
−3
d d b) 2
[cf( )] = c f( ) x
dx dx '
(cf)'= cf' ⎛ −3 ⎞ = ( 3 −2)' ( 3)( x−2 )' ( 3)( 2) − −1= 6 −3 = 6
⎜ 2 ⎟ 3
⎝ x ⎠ x
c) 2x
1 1
( 2x)'= 2( x)'= 2 =
2 x 2x
D Sum and Difference Rules Ex 4. Differentiate.
a) f (x) = 2 − 3x + 4x − 3x7
[ ( ) ± g ( )]f '( ) g '( )
f '(x) = (2 −3x + 4x −3x )'
d d d
[ ( ) ± g( )]= f( ) ± g(x) = (2)'+(−3x)'+(4x )'+(−3x )'
dx dx dx
( f ± g)'= f ' g' = 0 + (−3)(x)'+4(x )'−3(x )'
= −3+ 4(2)x −3(7)x 6
= −3+8x − 21x 6
∴ f '(x) = −3+8x − 21x 6
2.2 Derivative of Polynomial Functions
©2010 Iulia & Teodoru Gugoiu - Page 1 of 4 Calculus and Vectors – How to get an A+
1 2 4 x − 3 x
b) g(x) = − 2 + 5 c) h( ) =
x x x x
' ' '
⎛1 2 4 ⎞ −1 −2 −5 ⎛ x − 3 x⎞ ⎛ x 3x ⎞
g'( ) ⎜ − 2 + 5 ⎟ = ( − 2 + 4 )' h'( )= ⎜ ⎟ =⎜ − ⎟
⎝x x x ⎠ ⎝ x ⎠ ⎝ x x ⎠
− −2 −5
= ( )' ( 2x )' (4 )' 1−1 1−1 −1 −2 −1 −2
−2 −3 −6 = (x2 −x 3 )= (x 2 − x 3 )= (x 2 )− ( 3 )'
= − 1( ) ( 2)( 2)x + 4( 5)
1 4 20 ⎛− 1⎞ −−1 ⎛ − 2⎞ −2−1 ⎛− 1⎞ −3 ⎛ −2 ⎞ −5
= − + − = ⎜ ⎟x 2 −⎜ ⎟x 3 = ⎜ ⎟x 2 − ⎜ ⎟x 3
x 2 x3 x 6 ⎝ 2 ⎠ ⎝ 3 ⎠ ⎝ 2 ⎠ ⎝ 3 ⎠
∴ g x( )= − 1 + 4 − 20 −1 2 1 2
x2 x 3 x6 = 3 + 3 5 = − + 3 2
2 x 3 x 2 x x 3x x
1 2
∴ h'( )= − +
2x x 3 2
3 x x
E Tangent Line Ex 5. Find the equation of the tangent line at the point
To find the equation of the tangent line at the point 1
P(1,0)to the graph of y = f (x) = x − .
P(a.f (a)): x
1. Find derivative function f '(x). 1
f x ) = x +
2. Find the slope of the tangent line using: x2
m = f '(a) 2
m = f (1) = 2(1)+1/1 = 3
3. Use the slope-point formula to get the equation
of the tangent line: y −0 = 3(x −1)
∴ y = 3x −3
y − f (a) = m(x −a)
Ex 6. Find the equation of the tangent line of the 3 2
Ex 7. Find the points on the graph of y = f (x) = 2x −3x +1
slope m = 2 to the graph of y = f (x) = 2 x. Graph where the tangent line is horizontal.
the function and the tangent line.
1 1 f '(x) = 6x −6x = 6x(x −1)
f'( ) 2 =
2 x x f '(x) = m
1 1 1 m = 0 (horizontal tangent)
f'( )= 2 ⇒ = 2 ⇒ x= ⇒ y =2 =1 6x x − 1)= 0
x 4 4
P(1/4,1) ⇒ y− 1 2( x −1/4) x = 0 ⇒ y =1 ⇒ A(0,1)
∴ =y 2 x+ 1/2 x =1 ⇒ y= 0 ⇒ B (1,0)
The tangent line is horizontal at A(0,1)and B(1,0).
y = 2*x+1/2x) y
3 y
2
2
1
1
x
−2 −1 1 2 x
−3 −2 −1 1 2 3
−1 −1
−2
−2
−3
2.2 Derivative of Polynomial Functions
©2010 Iulia & Teodoru Gugoiu - Page 2 of 4 Calculus and Vectors – How to get an A+
F Normal Line Ex 8. Find the equation of the normal line to the curve
2
If m T is the slope of the tangent line, then slope of y = f (x) = x + at P(1,3) .
the normal line m is given by: x
N
1 2
m = − f '( )= − 2
N m x
T
m T = f(1) = −1
m = − 1 =1
N m
T
y −3 =1( −1)
∴ =y x + 2
Ex 9. Analyze the differentiability of each function. 2/3
b) y = f (x) = x
3 The function f is continuous over R .
a) y = f (x) = x 2 1
2 −1 2 − 2
The function f is continuous over R . f '(x) = x3 = x 3 =
1 1 2 3 3 3 x
1 −1 1 − 1
f'( ) = (x 3)'= x 3 = x 3 = f '(x)does not exist at x = 0 .
3 3 3 2
3 x The function f is not differentiable at x = 0 .
f '(x)does not exist at x = 0. Therefore the −
As x → 0 , f '(x) → −∞ .
function f is not differentiable at x = 0 .
As x → 0 +, f '(x) → ∞ .
As x → 0 , f '(x) → ∞ . The point O(0,0) is a infinite
The point O(0,0) is a cusp point.
slope point. 3 y
y
3
2
2
1
x
1
−3 −2 −1 1 2 3
x −1
−3 −2 −1 1 2 3
−1 −2
−3
−2
−3
y
c) y = f (x) =| x −3| 4
⎧ x −3, x ≥ 3 ⎧1, x > 3 3
f x) = ⎨ ⇒ f x( ) = ⎨ 2
3− ,x x < 3 −1, x < 3
⎩ ⎩ 1
lim f '(x) = −1 ≠ lim f '(x) =1 x
x→3− x→3 + −2 −1 1 2 3 4 5 6 7
−1
f 'does not exist at x = 3 . Therefore, f is not
−2
differentiable at x = 3 . The point P(3,0) is a corner −3
point (see the figure to the right side). −4
2.2 Derivative of Polynomial Functions
©2010 Iulia & Teodoru Gugoiu - Page 3 of 4 Calculus and Vectors – How to get an A+
H Differentiability for piece-wise defined Ex 10. Analyze the differentiability of each function at x = 1 .
functions ⎪ x2 , x ≤1
Let consider the piece-wise defined function: a) f( ) = ⎨
⎪ 2 , x >1
⎧ 1 (x), x < a ⎩
⎪
f (x) = ⎨, x = a
⎪ lim f (x) =1, lim f (x) = 2, f (1) =1
⎩ f2(x), x > a x→1− x→1 +
The function f is not continuous at x =1 . Therefore, the
The function f is differentiable at x = a if: function f is not differentiable at x =1 . See the figure
(a) the function is continuous at x = a below.
(b) f '(a) = f '(a) (the slope of the tangent line for
1 2 y
the left branch is equal to the slope of the 4
tangent line for the right branch). 3
2
1
x
−3 −2 −1 1 2 3
−1
⎪ x2 , x ≤1 ⎪ x2 , x ≤1
b) a) f x) = c) a) f x) =
⎪ x, x >1 ⎪ 2x−1, x >1
⎩ ⎩
lim f (x) =1, lim f (x) =1, f (1) =1 lim f (x) =1, lim f (x) =1, f (1) =1
x→1− x→1 + x→1− x→1 +
The function is continuous at x =1 . The function is continuous at x =1 .
⎧ 2 , x < 1 ⎧2 , x < 1
f'( ) = ⎨

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