Calculus and Vectors – How to get an A+
2.2 Derivative of Polynomial Functions
A Power Rule
Consider the power function: y = x n , x, n ∈ R .
Then:
Ex 1. For each case, differentiate.
a) (1)' = ( x 0 )' = 0 x 0−1 = 0
b) ( x )' = ( x1 )' = 1( x1−1 ) = 1( x 0 ) = 1
c) ( x 5 )' = 5( x 5−1 ) = 5 x 4
( x n )' = nx n −1
d n
x = nx n −1
dx
'
1
⎛1⎞
d) ⎜ ⎟ = ( x −1 )' = (−1) x −1−1 = − x − 2 = − 2
x
x
⎝ ⎠
'
−7
⎛ 1 ⎞
e) ⎜ 7 ⎟ = ( x −7 )' = (−7) x −7 −1 = −7 x −8 = 8
x
⎝x ⎠
Some useful specific cases:
(1)' = 0
1
f) ( x )' = ( x 2 )' =
( x)' = 1
( x )' =
1
1
2 x
g) (3 x )' = ( x 3 )' =
1
1
1
−2
1 2 −1 1 − 2 1 1
1
x
= x =
=
1
2
2
2
2 x
x2
1 3 −1 1 3
1
1
x
= x = 2 =
3 2
3
3
3 x
3x 3
3
h) ( 4 x 3 )' = ( x 4 )' =
−1
3
3 4 −1 3 4
3
3
x
= x = 1 =
4
4
4
4 x
4x 4
i) ( x π )' = πx π −1
B Constant Function Rule
Let consider a constant function: f ( x) = c, c ∈ R .
Then:
(c)' = 0
d
c=0
dx
Ex 2. Find each derivative function:
a) (−2)' = 0
C Constant Multiple Rule
Let consider g ( x) = cf ( x) . Then:
Ex 3. Differentiate each expression:
a) − 2x 3
(−2 x 3 )' = (−2)( x 3 )' = (−2)(3) x 3−1 = −6 x 2
−3
b) 2
x
[cf ( x)]' = cf ' ( x)
d
d
f ( x)
[cf ( x)] = c
dx
dx
(cf )' = cf '
b) ( π )' = 0
'
6
⎛ −3⎞
⎜ 2 ⎟ = (−3x − 2 )' = (−3)( x − 2 )' = (−3)(−2) x − 2−1 = 6 x −3 = 3
x
x
⎝
⎠
c)
2x
( 2 x )' = 2 ( x )' = 2
D Sum and Difference Rules
[ f ( x) ± g ( x)] = f ' ( x) ± g ' ( x)
d
d
d
f ( x) ±
[ f ( x) ± g ( x)] =
g ( x)
dx
dx
dx
( f ± g )' = f '± g '
1
2 x
=
Ex 4. Differentiate.
a) f ( x) = 2 − 3 x + 4 x 2 − 3x 7
f ' ( x) = (2 − 3x + 4 x 2 − 3x 7 )'
= (2)'+(−3x)'+(4 x 2 )'+(−3 x 7 )'
= 0 + (−3)( x)'+4( x 2 )'−3( x 7 )'
= −3 + 4(2) x − 3(7) x 6
= −3 + 8 x − 21x 6
∴ f ' ( x ) = −3 + 8 x − 21x 6
2.2 Derivative of Polynomial Functions
©2010 Iulia & Teodoru Gugoiu - Page 1 of 4
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2x
Calculus and Vectors – How to get an A+
b) g ( x) =
1
2
4
−
+
x x2 x5
'
'
2
4 ⎞
⎛1
g ' ( x) = ⎜ − 2 + 5 ⎟ = ( x −1 − 2 x − 2 + 4 x −5 )'
x
x
x ⎠
⎝
⎛ x −3 x ⎞ ⎛ x 3 x ⎞
⎟ =⎜
⎟
−
h' ( x ) = ⎜
⎜
⎟ ⎜ x
x
x ⎟⎠
⎝
⎠ ⎝
= ( x −1 )'+(−2 x − 2 )'+(4 x −5 )'
−2
x −3 x
x
c) h( x) =
1
−3
= −1( x ) + (−2)(−2) x + 4(−5) x
1
4
20
=− 2 + 3 − 6
x
x
x
1
4
20
∴ g ' ( x) = − 2 + 3 − 6
x
x
x
1
−1
−1
−
−2
1
'
−
−2
1
= ( x 2 − x 3 )' = ( x 2 − x 3 )' = ( x 2 )'−( x 3 )'
−6
−2
−1
−5
−3
⎛ − 1 ⎞ −1 ⎛ − 2 ⎞ 3 −1 ⎛ − 1 ⎞ 2 ⎛ − 2 ⎞ 3
= ⎜ ⎟x − ⎜
= ⎜ ⎟x 2 − ⎜
⎟x
⎟x
⎝ 3 ⎠
⎝ 2 ⎠
⎝ 3 ⎠
⎝ 2 ⎠
−1
2
1
2
=
+
=−
+
3 5
3
3
2 x x 3x x 2
2 x
3 x
∴ h' ( x ) = −
1
2x x
+
2
3
3x x 2
E Tangent Line
To find the equation of the tangent line at the point
P(a. f (a)) :
1. Find derivative function f ' ( x) .
2. Find the slope of the tangent line using:
m = f ' (a)
3. Use the slope-point formula to get the equation
of the tangent line:
y − f ( a ) = m( x − a )
Ex 5. Find the equation of the tangent line at the point
1
P(1,0) to the graph of y = f ( x) = x 2 − .
x
1
f ' ( x) = 2 x + 2
x
m = f (1) = 2(1) + 1 / 12 = 3
y − 0 = 3( x − 1)
∴ y = 3x − 3
Ex 6. Find the equation of the tangent line of the
Ex 7. Find the points on the graph of y = f ( x) = 2 x 3 − 3x 2 + 1
where the tangent line is horizontal.
slope m = 2 to the graph of y = f ( x) = 2 x . Graph
the function and the tangent line.
1
1
f ' ( x) = 2
=
2 x
x
f ' ( x) = 2 ⇒
P(1 / 4,1) ⇒
∴ y = 2 x + 1/ 2
1
1
=2 ⇒ x=
4
x
y − 1 = 2( x − 1 / 4)
⇒
f ' ( x) = 6 x 2 − 6 x = 6 x( x − 1)
f ' ( x) = m
m = 0 (horizontal tangent)
1
y=2
= 1 6 x( x − 1) = 0
4
x=0 ⇒
y =1 ⇒
A(0,1)
x = 1 ⇒ y = 0 ⇒ B(1,0)
The tangent line is horizontal at A(0,1) and B (1,0) .
y
y = 2*sqrt(x)
y = 2*x+1/2
3
y
2
2
1
1
x
−2
−1
1
x
2
−3
−2
−1
−1
1
−1
−2
−2
−3
2.2 Derivative of Polynomial Functions
©2010 Iulia & Teodoru Gugoiu - Page 2 of 4
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Calculus and Vectors – How to get an A+
F Normal Line
If mT is the slope of the tangent line, then slope of
the normal line m N is given by:
1
mN = −
mT
Ex 8. Find the equation of the normal line to the curve
2
y = f ( x) = x + at P (1,3) .
x
2
f ' ( x) = 1 − 2
x
mT = f (1) = −1
mN = −
1
=1
mT
y − 3 = 1( x − 1)
∴y = x+2
Ex 9. Analyze the differentiability of each function.
a) y = f ( x) = 3 x
The function f is continuous over R .
1
1
2
1 3 −1 1 − 3
3
f ' ( x) = ( x )' = x
= x =
3
3
2
1
3
3 x2
f ' ( x) does not exist at x = 0 . Therefore the
function f is not differentiable at x = 0 .
As x → 0 , f ' ( x) → ∞ . The point O(0,0) is a infinite
slope point.
3
b) y = f ( x) = x 2 / 3
The function f is continuous over R .
1
2 −1 2 −
2
f ' ( x) = x 3 = x 3 =
3
3
3
3 x
f ' ( x) does not exist at x = 0 .
The 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 cusp point.
3
y
y
2
2
1
x
1
−3
−2
−1
1
x
−3
−2
−1
1
2
2
3
−1
3
−2
−1
−3
−2
−3
c) y = f ( x) =| x − 3 |
⎧ x − 3,
f ( x) = ⎨
⎩3 − x,
x≥3
⎧1, x > 3
⇒ f ' ( x) = ⎨
x<3
⎩− 1, x < 3
lim− f ' ( x) = −1 ≠
x→3
y
4
lim+ f ' ( x) = 1
3
2
1
x
x→3
−2
−1
f ' does not exist at x = 3 . Therefore, f is not
differentiable at x = 3 . The point P(3,0) is a corner
point (see the figure to the right side).
2.2 Derivative of Polynomial Functions
©2010 Iulia & Teodoru Gugoiu - Page 3 of 4
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−1
−2
−3
−4
2
3
4
5
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Calculus and Vectors – How to get an A+
H Differentiability for piece-wise defined
functions
Let consider the piece-wise defined function:
⎧ f1 ( x), x < a
⎪
f ( x) = ⎨c, x = a
⎪ f ( x), x > a
⎩ 2
The function f is differentiable at x = a if:
(a) the function is continuous at x = a
(b) f1 ' (a) = f 2 ' (a) (the slope of the tangent line for
the left branch is equal to the slope of the
tangent line for the right branch).
Ex 10. Analyze the differentiability of each function at x = 1 .
⎧⎪ x 2 , x ≤ 1
a) f ( x) = ⎨
⎪⎩2 x, x > 1
lim f ( x ) = 1,
x→1−
lim f ( x) = 2,
x→1+
f (1) = 1
The function f is not continuous at x = 1 . Therefore, the
function f is not differentiable at x = 1 . See the figure
below.
y
4
3
2
1
x
−3
−2
−1
1
2
3
−1
⎧⎪ x 2 , x ≤ 1
b) a) f ( x) = ⎨
⎪⎩ x, x > 1
lim f ( x) = 1,
⎧⎪ x 2 , x ≤ 1
c) a) f ( x) = ⎨
⎪⎩2 x − 1, x > 1
lim f ( x ) = 1,
x→1−
x→1+
f (1) = 1
lim f ( x) = 1,
x→1−
The function is continuous at x = 1 .
⎧2 x , x < 1
f ' ( x) = ⎨
⎩1, x > 1
lim f ' ( x) = 2,
f (1) = 1
The function is continuous at x = 1 .
⎧ 2 x, x < 1
f ' ( x) = ⎨
⎩2, x > 1
lim f ( x) = 1
x→1−
lim f ( x) = 1,
x→1+
lim f ' ( x) = 2,
x→1+
x→1−
lim f ( x) = 2
x→1+
The function f is not differentiable at x = 1 (corner The function f is differentiable at x = 1 . See the figure
point). See the figure below.
below.
5
y
y
4
4
3
3
2
2
1
1
x
−3
−2
−1
1
2
3
x
−3
−2
−1
1
2
3
−1
−1
−2
Reading: Nelson Textbook, Pages 76-81
Homework: Nelson Textbook: Page 82 #2cdf, 3ce, 4aef, 5b, 6b, 7a, 8b, 9b, 11, 14, 17, 20, 25iii, 28
2.2 Derivative of Polynomial Functions
©2010 Iulia & Teodoru Gugoiu - Page 4 of 4