22_Derivative_of_Polynomial_Functions.pdf

Calculus and Vectors – How to get an A+

2.2 Derivative of Polynomial Functions ©2010 Iulia & Teodoru Gugoiu - of 4

2.2 Derivative of Polynomial Functions A Power Rule Consider the power function: Rnxxy n ∈= ,, . Then:

1)'( −= nn nxx 1−= nn nxx

dxd

Some useful specific cases:

xx

x

21)'(

1)'(0)'1(

=

==

Ex 1. For each case, differentiate. a) 00)'()'1( 100 === −xx

b) 1)(1)(1)'()'( 0111 ==== − xxxx

c) 4155 5)(5)'( xxx == −

d) 22111

' 1)1()'(1x

xxxx

−=−=−==⎟⎠⎞

⎜⎝⎛ −−−−

e) 88177

'

777)7()'(1x

xxxx

−=−=−==⎟

⎠

⎞⎜⎝

⎛ −−−−

f) x

x

xxxx211

21

21

21)'()'(

21

211

21

21

=====−−

g) 3 2

32

321

31

31

3

3

1

3

131

31)'()'(

xx

xxxx =====−

−

h) 4

41

411

43

43

4 3

43

4

343

43)'()'(

xx

xxxx =====−

−

i) 1)'( −= ππ πxx

B Constant Function Rule Let consider a constant function: Rccxf ∈= ,)( . Then:

0)'( =c

0=cdxd

Ex 2. Find each derivative function: a) 0)'2( =−

b) 0)'( =π

C Constant Multiple Rule Let consider )()( xcfxg = . Then:

')'(

)()]([

)(')]'([

cfcf

xfdxdcxcf

dxd

xcfxcf

=

=

=

Ex 3. Differentiate each expression: a) 32x−

21333 6)3)(2()')(2()'2( xxxx −=−=−=− −

b) 23x−

331222

'

266)2)(3()')(3()'3(3x

xxxxx

==−−=−=−=⎟⎠

⎞⎜⎝

⎛ − −−−−−

c) x2

xxxx

21

212)'(2)'2( ===

D Sum and Difference Rules

'')'(

)()()]()([

)(')(')]()([

gfgf

xgdxdxf

dxdxgxf

dxd

xgxfxgxf

±=±

±=±

±=±

Ex 4. Differentiate. a) 72 3432)( xxxxf −+−=

6

6

6

72

72

72

2183)('

2183

)7(3)2(43

)'(3)'(4)')(3(0

)'3()'4()'3()'2(

)'3432()('

xxxf

xx

xx

xxx

xxx

xxxxf

−+−=∴

−+−=

−+−=

−+−+=

−++−+=

−+−=



b) 52421)(xxx

xg +−=

632

632

632

521

521'

52

2041)('

2041)5(4)2)(2()(1

)'4()'2()'(

)'42(421)('

xxxxg

xxx

xxx

xxx

xxxxxx

xg

−+−=∴

−+−=

−+−−+−=

+−+=

+−=⎟⎠

⎞⎜⎝

⎛ +−=

−−−

−−−

−−−

c) xxxxh

3)( −=

3 2

3 23 53

35

231

321

21

32

21

32

211

311

21

'3'3

3

221)('

3

221

3

2

2

132

21

32

21

)'()'()'()'(

)('

xxxxxh

xxxxxx

xxxx

xxxxxx

xx

xx

xxxxh

+−=∴

+−=+−

=

⎟⎠⎞

⎜⎝⎛ −−⎟

⎠⎞

⎜⎝⎛ −=⎟

⎠⎞

⎜⎝⎛ −−⎟

⎠⎞

⎜⎝⎛ −=

−=−=−=

⎟⎟⎠

⎞⎜⎜⎝

⎛−=⎟

⎟⎠

⎞⎜⎜⎝

⎛ −=

−−−

−−

−

−−

−−−−

E Tangent Line To find the equation of the tangent line at the point

))(.( afaP : 1. Find derivative function )(' xf . 2. Find the slope of the tangent line using:

)(' afm = 3. Use the slope-point formula to get the equation of the tangent line:

)()( axmafy −=−

Ex 5. Find the equation of the tangent line at the point

)0,1(P to the graph of x

xxfy 1)( 2 −== .

33)1(30

31/1)1(2)1(

12)('

2

2

−=∴−=−

=+==

+=

xyxy

fmx

xxf

Ex 6. Find the equation of the tangent line of the slope 2=m to the graph of xxfy 2)( == . Graph the function and the tangent line.

2/12)4/1(21)1,4/1(

1412

41212)('

1212)('

+=∴−=−⇒

==⇒=⇒=⇒=

==

xyxyP

yxx

xf

xxxf

−2 −1 1 2

−2

−1

1

2

x

yy = 2*sqrt(x)

y = 2*x+1/2

Ex 7. Find the points on the graph of 132)( 23 +−== xxxfy where the tangent line is horizontal.

mxfxxxxxf

=−=−=

)(')1(666)(' 2

0=m (horizontal tangent)

)0,1(01)1,0(10

0)1(6

ByxAyx

xx

⇒=⇒=⇒=⇒=

=−

The tangent line is horizontal at )1,0(A and )0,1(B .

−3 −2 −1 1 2 3

−3

−2

−1

1

2

3

x

y



F Normal Line If Tm is the slope of the tangent line, then slope of the normal line Nm is given by:

TN mm 1

−=

Ex 8. Find the equation of the normal line to the curve

xxxfy 2)( +== at )3,1(P .

2)1(13

111)1(

21)(' 2

+=∴−=−

=−=

−==

−=

xyxym

m

fmx

xf

TN

T

Ex 9. Analyze the differentiability of each function. a) 3)( xxfy == The function f is continuous over R .

3 2321

31

31

3

131

31)'()('

xxxxxf ====−−

)(' xf does not exist at 0=x . Therefore the function f is not differentiable at 0=x . As 0→x , ∞→)(' xf . The point )0,0(O is a infinite slope point.

−3 −2 −1 1 2 3

−3

−2

−1

1

2

3

x

y

b) 3/2)( xxfy == The function f is continuous over R .

3311

32

32

32

32)('

xxxxf ===−−

)(' xf does not exist at 0=x . The function f is not differentiable at 0=x .

As −→ 0x , −∞→)(' xf .

As +→ 0x , ∞→)(' xf . The point )0,0(O is a cusp point.

−3 −2 −1 1 2 3

−3

−2

−1

1

2

3

x

y

c) |3|)( −== xxfy

⎩⎨⎧

<−≥−

=3,33,3

)(xxxx

xf⎩⎨⎧

<−>

=⇒3,13,1

)('xx

xf

1)('lim1)('lim33

=≠−=+− →→

xfxfxx

'f does not exist at 3=x . Therefore, f is not differentiable at 3=x . The point )0,3(P is a corner point (see the figure to the right side).

−2 −1 1 2 3 4 5 6 7

−4

−3

−2

−1

1

2

3

4

x

y



H Differentiability for piece-wise defined functions Let consider the piece-wise defined function:

⎪⎩

⎪⎨

⎧

>=

<=

axxfaxc

axxfxf

),(,),(

)(

2

1

The function f is differentiable at ax = if: (a) the function is continuous at ax = (b) )(')(' 21 afaf = (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 1=x .

a) ⎪⎩

⎪⎨⎧

>≤

=1,21,)(

2

xxxxxf

1)1(,2)(lim,1)(lim

11===

+− →→fxfxf

xx

The function f is not continuous at 1=x . Therefore, the function f is not differentiable at 1=x . See the figure below.

−3 −2 −1 1 2 3

−1

1

2

3

4

x

y

b) a) ⎪⎩

⎪⎨⎧

>≤

=1,1,)(

2

xxxxxf

1)1(,1)(lim,1)(lim

11===

+− →→fxfxf

xx

The function is continuous at 1=x .

⎩⎨⎧

><

=1,11,2

)('xxx

xf

1)(lim,2)('lim11

==+− →→

xfxfxx

The function f is not differentiable at 1=x (corner point). See the figure below.

−3 −2 −1 1 2 3

−1

1

2

3

4

x

y

c) a) ⎪⎩

⎪⎨⎧

>−≤

=1,12

1,)(2

xxxxxf

1)1(,1)(lim,1)(lim

11===

+− →→fxfxf

xx

The function is continuous at 1=x .

⎩⎨⎧

><

=1,21,2

)('xxx

xf

2)(lim,2)('lim11

==+− →→

xfxfxx

The function f is differentiable at 1=x . See the figure below.

−3 −2 −1 1 2 3

−2

−1

1

2

3

4

5

x

y

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

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