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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)
Ex. 1: Find the derivative of each function.
(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)
You Try: Find the derivative of each function.
(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)
Ex. 4: Find the derivative of each function.
(a)
(b)
(c)
(d)
Ex. 4: Find the derivative of each function.
(a)
(b)
(c)
(d)
y=−4x2
5
f (t)=6t3
f (x)=5 x23
y=2x
You Try: Find the derivative of each function.
(a)
(b)
(c)
You Try: Find the derivative of each function.
(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)
Ex. 5: Find the derivative of each function.
(a)
(b)
Ex. 5: Find the derivative of each function.
(a)
(b)
f (x)=4x3 −2x+9
g(t)=−t3
6−t2 +6
You Try: Find the derivative of each function.
(a)
(b)
You Try: Find the derivative of each function.
(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
Ex. 6: Find the derivative of each function.
(a)
(b)
Ex. 6: Find the derivative of each function.
(a)
(b)
y=6cosx
y=3x2 −sinx
You Try: Find the derivative of each function.
(a)
(b)
You Try: Find the derivative of each function.
(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