2.2 Differentiation Rules for Constant Multiples, Sums, Powers, Sines, and Cosines

Constant Rule: The derivative of a constant is zero.

0][ =cdxd

Find the derivatives of:

5)('

0)(

7

−==

=

tsxf

y 0'=y0)(' =xf

0)(" =ts

Power Rule: If n is a rational number, then

1][ −= nn nxxdxd

Find the derivatives of:

1006416)(

)(

1

)(

2

2

3

++−=

=

=

=

ttts

xxgx

y

xxf23)(' xxf =

rewritten as 2−=x 32' −−= xy3

2

x

−=

1)(' =xg

6432)()(' +−== ttvts

5

4)(

2

2ttf

xy

=

=

Differentiate:

12 −= x 22 2

2x

xdx

dy−=−= −

5

8)('

ttf =

[ ]=

=⎥⎦⎤

⎢⎣

⎡−

xdx

d

x

dx

d

π3

2

3

2

3−

π3

Sum and Difference Rules

[ ]

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

)(')(')()(

xgxfxgxfdx

d

xgxfxgxfdx

d

−=−

+=+

xxx

xg 232

)( 34

−+−= 292)(' 23 −+−= xxxg

Differentiate:

3 22

1

2

xy

xy

=

= 21

2x=x

xxy1

2

12' 2

12

1==⋅=

−−

32

2

1 −= x 3

5

3

2

2

1 −⎟⎠

⎞⎜⎝

⎛−= xdxdy

35

3

1

x−=

Derivatives of Sine and Cosine

[ ] [ ] xxdx

dxx

dx

dsincoscossin −==

xxy

xy

cos

sin3

+== xy cos3'=

xy sin1' −=

Find the slope and equation of the tangent lineof the graph of y = 2 cos x at the point .1,

3⎟⎠

⎞⎜⎝

⎛π

f’(x) = -2sin x

=−=⎟⎠

⎞⎜⎝

⎛3

sin23

'@ππ

f 32

32 −=⎟⎟

⎠

⎞⎜⎜⎝

⎛−

Therefore, the equation of the tangent line is:

⎟⎠

⎞⎜⎝

⎛ −−=−3

31π

xy

Day 1

The average rate of change in distance withrespect to time is given by…

change in distancechange in time t

s

ΔΔ

= Also known asaverage velocity

Ex. If a free-falling object is dropped from aheight of 100 feet, its height s at time t is givenby the position function s = -16t2 + 100, wheres is measured in feet and t is measured in seconds.Find the average rate of change of the height overthe following intervals.

a. [1, 2] b. [1, 1.5] c. [1, 1.1]

a. =t

s

Δ

Δ =−−

12

8436sec/48

1

48ft−=

−

b. =t

s

Δ

Δ =−

−15.1

8464sec/40

5.

20ft−=

−

c. =t

s

Δ

Δ =−−11.1

8464.80sec/6.33

1.

36.3ft−=

−

At time t = 0, a diver jumps from a diving board that is 32 feet above the water. The position of the diver is given by

321616)( 2 ++−= tttswhere s is measured in feet and t in seconds.

a. When does the diver hit the water?b. What is the diver’s velocity at impact?

To find the time at which the diver hits the water,we let s(t) = 0 and solve for t.

3216160 2 ++−= tt

( )2160 2 −−−= tt

( )( )21160 −+−= tt

t = -1 or 2

-1 doesn’t make sense, so the diver hits at 2 seconds.

The velocity at time t is given by the derivative.

s’(t) = v(t) = -32t + 16

@ t = 2 seconds, s’(2) = -48 ft/sec.

The negative gives the direction, which in this case is down.