basic differentiation rules and rates of change (2.2) october 12th, 2011

Basic Differentiation rules and rates Basic Differentiation rules and rates

ofof change (2.2)change (2.2) Basic Differentiation rules and rates Basic Differentiation rules and rates

ofof change (2.2)change (2.2) October 12th, 2011October 12th, 2011

I. The constant ruleI. The constant rule

Thm. 2.2: The Constant Rule: The

derivative of a constant function is 0.

So, if c is a real number, then

.

Thm. 2.2: The Constant Rule: The

derivative of a constant function is 0.

So, if c is a real number, then

.

d

dx[c]=0

II. THe power ruleII. THe power ruleThm. 2.3: The Power Rule: If n is a rational number, then

the function is differentiable and

.

For f to be differentiable at x=0, n must be a number such

that is defined on an interval containing

0.

*When n=1, .

Therefore, .

Thm. 2.3: The Power Rule: If n is a rational number, then

the function is differentiable and

.

For f to be differentiable at x=0, n must be a number such

that is defined on an interval containing

0.

*When n=1, .

Therefore, .

f (x)=xn

d

dx[xn ]=nxn−1

xn−1

d

dx[x1]=1x1−1 =x0 =1

d

dx[x]=1

A. Using the power ruleA. Using the power rule

Ex. 1: Find the derivative of each function.

(a)

(b)

(c)

(d)

(e)

(f)


(a)

(b)

(c)

(d)

(e)

(f)

f (x)=x5

g(x)= x

y=x

h(x)=4

k(x)=cπ 3

y=1t3

You Try: Find the derivative of each function.

(a)

(b)

(c)

(d)


(a)

(b)

(c)

(d)

y=x2

f (t)=1t5

g(x)= x4

h(x)=e

B. Finding the slope of a graphB. Finding the slope of a graph

Ex. 2: Find the slope of the graph of

when

(a) x = -2

(b) x = 0

(c) x = 2

Ex. 2: Find the slope of the graph of

when

(a) x = -2

(b) x = 0

(c) x = 2

f (x)=x3

You Try: Find the slope of the graph of

when

(a) x = -1

(b) x = 1

You Try: Find the slope of the graph of

when

(a) x = -1

(b) x = 1

y=1x2

C. finding an equation of a tangent line C. finding an equation of a tangent line

Ex. 3: Find an equation of the tangent line to the graph of

when x = 2.

Ex. 3: Find an equation of the tangent line to the graph of

when x = 2.f (x)=x4

You Try: Find an equation of the tangent line to the graph

of

when x = -3.

You Try: Find an equation of the tangent line to the graph

of

when x = -3.f (x)=x2

III. the constant multiple ruleIII. the constant multiple rule

Thm. 2.4: The Constant Multiple Rule:

If f is a differentiable function and c is

a real number, then cf is also

differentiable and

.

Thm. 2.4: The Constant Multiple Rule:

If f is a differentiable function and c is

a real number, then cf is also

differentiable and

.

d

dx[cf (x)]=cf '(x)


(a)

(b)

(c)

(d)


(a)

(b)

(c)

(d)

y=−4x2

5

f (t)=6t3

f (x)=5 x23

y=2x


(a)

(b)

(c)


(a)

(b)

(c)

g(t)=3t3

5

y=5x

h(x)=4

3x4

IV. The sum and difference rulesIV. The sum and difference rules

Thm. 2.5: The Sum and Difference

Rules: The sum or difference of two

differentiable functions f and g is also

differentiable and

Thm. 2.5: The Sum and Difference

Rules: The sum or difference of two

differentiable functions f and g is also

differentiable andd

dx[ f (x)+g(x)] = f '(x)+g'(x)

d

dx[ f (x)−g(x)] = f '(x)−g'(x)


(a)

(b)


(a)

(b)

f (x)=4x3 −2x+9

g(t)=−t3

6−t2 +6


(a)

(b)


(a)

(b)

y=4x3 − x

f (t)=−3t5 +2t3

−6

V. Derivatives of Sine and CosineV. Derivatives of Sine and Cosine

Thm. 2.6: Derivatives of Sine and

Cosine Functions:

Thm. 2.6: Derivatives of Sine and

Cosine Functions: d

dx[sin x]=cosx

d

dx[cos x]=−sinx


(a)

(b)


(a)

(b)

y=6cosx

y=3x2 −sinx


(a)

(b)


(a)

(b)

y=2cosx

5y=5x3 −2x+ 3sinx

VI. Rates of changeVI. Rates of change*The function s is called the position function and gives

the position of an object with respect to the origin as a

function of time t. The change in position over a period of

time is given by

and we know rate = distance/time, so we know the

average velocity is

average velocity = (change in distance)/(change in time)

= .

*The function s is called the position function and gives

the position of an object with respect to the origin as a

function of time t. The change in position over a period of

time is given by

and we know rate = distance/time, so we know the

average velocity is

average velocity = (change in distance)/(change in time)

= .

Δt

Δs = s(t + Δt) − s(t)

ΔsΔt

A. finding average velocity of a falling objectA. finding average velocity of a falling object

Ex. 7: Given the position function

, find the average velocity over the interval [2, 2.1].

Ex. 7: Given the position function

, find the average velocity over the interval [2, 2.1].

s(t)=t2 −3

You Try: Given the position function s(t)= sin t, find the

average velocity over the interval [0, ].

You Try: Given the position function s(t)= sin t, find the

average velocity over the interval [0, ].π6

*The instantaneous velocity of an object at time t is

So, the velocity function v(t) is given by the derivative of

the position function s(t).

*The speed of an object is given by the absolute value of

the velocity (velocity has direction, but speed cannot be

negative).

*The instantaneous velocity of an object at time t is

So, the velocity function v(t) is given by the derivative of

the position function s(t).

*The speed of an object is given by the absolute value of

the velocity (velocity has direction, but speed cannot be

negative).

v(t)=limΔt→ 0

s(t+Δt)−s(t)Δt

=s'(t)

*The position of a free-falling object (neglecting air

resistance) under the influence of gravity is given by

where is the initial height, is the initial

velocity, and g is the acceleration due to gravity.

(On Earth, feet per second per second or

meters per second per second).

*The position of a free-falling object (neglecting air

resistance) under the influence of gravity is given by

where is the initial height, is the initial

velocity, and g is the acceleration due to gravity.

(On Earth, feet per second per second or

meters per second per second).

s(t)=12gt2 +v0t+ s0 ,

s0 v0

g≈−32 g≈−9.8

B. Using the derivative to find velocityB. Using the derivative to find velocity

Ex. 8: At time t = 0, a person jumps off a cliff that is 980

meters above the ground. The position of the person is

given by

where s is measured in meters and t is in seconds.

(a) When does the person hit the ground?

(b) What is the person’s velocity at impact?

Ex. 8: At time t = 0, a person jumps off a cliff that is 980

meters above the ground. The position of the person is

given by

where s is measured in meters and t is in seconds.

(a) When does the person hit the ground?

(b) What is the person’s velocity at impact?

s(t)=−4.9t2 +2t+980

basic differentiation rules and rates of change (2.2) october 12th, 2011

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