Download Math 111 – Calculus I

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Math 111 – Calculus I.
Week Number Seven Notes
Fall 2002
I.
Derivatives of Trigonometric Functions
We’ll begin with a statement of the derivatives of each of the trigonometric functions.
Theorem 7.1:
(i) (sin(x))'  cos(x)
(iv) (cot(x))'  - csc 2 (x)
(ii) (cos(x))'  sin(x)
(v) (sec(x))'  sec(x)tan( x)
(iii) (tan(x))'  sec 2 ( x)
(vi) (csc(x))'  - csc(x)cot( x)
Please note the following.
(i)
The derivatives of (iii) through (vi) can be derived using the quotient rule and
the derivative formulas for sin(x) and cos(x) (I will prove that the definition of
sec(x) is sec(x)tan(x) shortly.
(ii)
The derivative of cos(x) can be derived relatively “easily” using elements of
the argument that the derivative of sin(x) is cos(x).
(iii)
The key to proving that the derivative of sin(x) is cos(x) is based upon a result
derived using geometric principle. It is as follows.
For any real number x,
cos(x) < sin(x)/x < 1.
Using this result and the pinching theorem for limits, the following IMPORTANT
LIMIT FOLLOWS (see section 3.4, p. 220 of the textbook for details).
lim
x 0
sin(x)
1
x
Now, I’ll argue the following result assuming derivative formulas (i) and (ii) in theorem
7.1 and the quotient rule.
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Example 7.2: d(sec(x))/dx = sec(x)tan(x)
Proof:
d(sec(x)) d(1/cos(x) )


dx
dx
cos(x)(
d(1)
d(cos(x))
)  1(
)
sin(x)
1 sin(x)
dx
dx


 sec( x)tan(x)
2
2
cos (x)
cos ( x) cos(x) cos(x)
Now, let’s do a few more examples.
Example 7.3: Differentiate the following functions below.
(a) f(x)  xcos(x)
(b) g(x) 
tan(x)
x
(c) h(x) 
x
sin(x)  cos(x)
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Example 7.4(section 3.4 of textbook, problem 29 on p. 224): A mass on a spring vibrates
horizontally on a smooth level surface. Its equation of motion is x(t) = 8sin(t) where t
represents the time in seconds and x is the displacement from rest position in centimeters.
(a) Find equations for the instantaneous velocity and instantaneous acceleration of the
mass at time t.
(b) Find the displacement, instantaneous velocity, and instantaneous acceleration of
the mass at time t =  Interpret the results.
II.
The Chain Rule
The most common method of constructing new functions from old ones is to use function
composition . Unfortunately, we do not have a rule to differentiate a composition of
functions f and g. This is dealt with by the Chain Rule. It’s statement is as follows.
Theorem 7.5(THE CHAIN RULE): Assume f and g are differentiable and F(x) = f(g(x)).
Then,
F' (x)  f' (g(x))g' (x)
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Using Leibniz style notation,
If y  f(u) and u  g(x), then
dy dy du

dx du dx
A proof of the Chain Rule is given in the textbook on p. 232. Let’s do a few examples to
get the feeling on how the Chain Rule works.
Example 7.6: Compute the derivative of the following function.
f(x)  x 4  2x 2  1
Solution: Assume g(x) = x4 + 2x2 + 1 and h(x) = x1/2. Then, by the Chain Rule,
f' (x)  (h(g(x))'  h' (g(x))g' (x)
Now, h' (x) 
1 -1/2
1
x and g' (x)  4x 3  4x. Thus, f' (x)  ( x 4  2x 2  1) -1/2 (4x 3  4x).
2
2
Here are two more examples.
Example 7.7: Compute the derivatives of the following functions.
(a) f(x) = (cos(x))3
(b) g(x) = sin(sin(sin(x)))
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Example 7.8(Section 3.5, problem 62, p. 235 – an example of related rates): Assume air
is being pumped into a spherical weather balloon. Assume at any time t, V(t) represents
the volume of the balloon and r(t) represents the radius of the balloon. Find dV/dt in
terms of V, r, and dr/dt.
III.
Implicit Differentiation
So far, we have learned to differentiate functions. However, suppose we want to find the
slope of the tangent line to curve at a given value in which y is not a function of x (or
equivalently x is not a function of y – for example the unit circle). One option, is to solve
for y in terms of x on a restricted domain. However, this is often times EXTREMELY
DIFFICULT. However, we do not need to do this. We just assume that y is a function of
x on a sufficiently small open interval about x, differentiate the relation, and solve for
dy/dx in terms of x and y. This technique is called IMPLICIT DIFFERENTIATION.
Let’s look at two examples that illustrates this technique.
Example 7.9(similar to example 2, p. 240 of the textbook):
(a) Find the equation of the tangent line to the curve x3 + y3 = 6xy at the point (3,3).
(b) At what points on the curve (if any) does a horizontal tangent line exist? a
vertical tangent line?
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Example 7.10: Use implicit differentiation to find d(arctan(x))/dx.
Non Hand-In Homework Problems Associated with Week #7 Notes
Sections 3.4 – 3.6
Section 3.4: 1-12,15,18,20,23,25,29,31,37,40
Section 3.5: 1-31, 33,36,51,53,59,60
Section 3.6: 1-15,18, 27,31,33, 49,50
Read Sections 3.3 and 3.7 of the textbook