Chat with us, powered by LiveChat Calculus | acewriters
+1(978)310-4246 credencewriters@gmail.com
Select Page

11 Lecture need to Solve Problems On Calculus 12.2, 2.3, 2.4, 2.5, 2.6, 2.7, 2.8, 2.9, 3.1, 3.2, 3.3There are 72 problems in total
2.2.pdf

2.3.pdf

2.4.pdf

2.5.pdf

2.6.pdf

Unformatted Attachment Preview

Section 2.2 / The derivative as a function
Overview
Last section, we defined the derivative (slope) of a function at a point. It
takes a little bit of computation to find a slope. Suppose we want to find the
slope of y = x3 at x = 0, 1, 2, 4, 7, 8, 12, 13, and 18? Rather than go through
the process of computing the limit for each of these numbers, we can compute
the limit at an arbitrary (unspecified) number, and then plug in whichever
specific number we like later. This leads to the definition of the derivative of
a function, which is itself a function.
Details
def: The derivative of the function f (x) is the function defined by
f (x + h) − f (x)
.
h→0
h
f 0 (x) = lim
rk: The domain of f 0 (x) is the set of all x for which the above limit exists.
Sometimes this is the same as the domain of f (x), but sometimes it is smaller.
discuss:
Given the graph of y = f (x), can you sketch the graph of
0
y = f (x)? How about vice versa?
ex: Find the derivative of f (x) =

x. Graph f and f 0 on the same axes.
ex: Find the derivative of
f (x) =
1−x
.
2+x
def: As stated in section 2.1, f is called differentiable at a if f 0 (a) exists.
Otherwise we say f is not differentiable at a.
def: We say f is differentiable on (a, b) if f is differentiable at each number
in the interval (a, b).
th: If f is differentiable at a, then f is continuous at a.
1
discuss: What does the graph of a function look like at a point where the
function is not differentiable?
discuss: Higher order derivatives – rate of change of the slope. Since the
derivative is just another function, you can take the derivative of a derivative.
The second derivative of position with respect to time is called acceleration.
discuss: The units of a derivative are the units of the function of the dependant variable per unit of the independent variable. For example, if position
is measured in feet and time is measured in seconds, then the first derivative
(velocity) is measured in feet per second. The second derivative would be
measured in feet per second per second.
Key vocab, techniques
vocab: derivative (of a function), differentiable (at a point or on an interval), differentiable / not differentiable, acceleration
techniques: Given a function, find its derivative (by using the limit definition of the derivative). Find the equation of the tangent line to a function at
a given number. Given the graph of a function, determine the graph of its
derivative.
Problems
1. Let f (x) =

1 − x.
(a) Use the limit definition of the derivative to find f 0 (x). Do not use
shortcut formulas.
(b) Give the domain of f (x) and f 0 (x).
(c) Find the equation of the tangent line to y = f (x) at x = −3.

2. Let f (x) = 2x + 1.
(a) Use the limit definition of the derivative to find f 0 (x). Do not use
shortcut formulas.
(b) Give the domain of f (x) and f 0 (x).
(c) Find the equation of the tangent line to y = f (x) at x = 4.
2
3. Let f (x) =

x + 5.
(a) Use the limit definition of a derivative to find f 0 (x). Please do not use
shortcut formulas.
(b) Find the equation of the tangent line to y = f (x) at x = −1.

4. Let f (x) = 1 − 2x.
(a) Use the limit definition of the derivative to find f 0 (x). Do not use
shortcut formulas.
(b) Give the domain of f (x) and f 0 (x).
(c) Find the equation of the tangent line to y = f (x) at x = −4.
5. Let f (x) =
1
.
x2 +1
(a) Use the limit definition of a derivative to find f 0 (x). Do not use shortcut
formulas.
(b) Give the domain of f (x) and f 0 (x).
(c) Find the equation of the tangent line to y = f (x) at x = −1.
6. Let f (x) =
√1 .
x
(a) Find f 0 (x) using the limit definition of the derivative. Do not use
shortcut formulas.
(b) Find the equation of the line tangent to y = f (x) at x = 4.
7. Use the limit definition of a derivative to find the derivative:
 
d
x3
dx
8. Use the limit definition of a derivative to find the derivative:
d 2
(x − x)
dx
3
Section 2.3 / Differentiation formulas
Overview
Having introduced the concept of a derivative of a function, we’d like to be
able to compute derivatives as easily as possible. Remember the limit laws?
Noticing some facts about limits of some simple functions, and how those
functions combine, let us take limits of more complicated functions. We’ll
do the same thing for derivatives in this section.
Details
fact:

d
(c)
dx
= 0 for any constant c

d
(x)
dx
=1

d
(cf )
dx
d
= c dx
f for any constant c

d
(xr )
dx
= rxr−1 for any real number r (the power rule)

d
(f
dx

d
(f g)
dx

 0
f
g
± g) =
=
d
f
dx
±
d
g
dx
(the addition rule)
= f 0 g + g 0 f (the product rule)
f 0 g−g 0 f
g2
(the quotient rule)
ex: Find the derivatives of:
• f (x) = x4 − 3×2 + 4

• x x

• x x+1

1−x
2+x

x2 +x−2
x3 +6
Key vocab, techniques
vocab: power rule, product rule, quotient rule
techniques: Apply the differentiation formulas to find derivatives of functions.
Problems
1
1. Find the derivative: f (x) =
2. Using the facts that
where
d
dx
1
.
x2 +1
sin x = cos x and
d
dx
cos x = − sin x, find f 0 (x)

4
3
x2 tan x − x3 x4
f (x) =
.
sin(2x)
3. Find the equation of the line tangent to f (x) =
x
1+x2
at x = 1.

4.
Find the point on the graph y = x x with slope 3. Then find the
equation of the line tangent to the graph at that point.
5. Let f (x) = 13 x3 + 3×2 + 10x − 7.
(a) Show that f 0 (x) > 0 for every real number x.
(b) Argue that the graph y = f (x) is therefore always increasing.
(c) Conclude that f (x) has exactly one real zero.
2
Section 2.4 / Derivatives of trig functions
Overview
We find the derivatives of the six trigonometric functions.
Details
discuss: Now that we know what a graph of a function ought to look
like next to the graph of its derivative, let’s draw the graphs y = sin x and
y = cos x. Let’s then try hard to notice some things.
fact:

d
dx
sin x = cos x

d
dx
csc x = − csc x cot x

d
dx
cos x = − sin x

d
dx
sec x = sec x tan x

d
dx
tan x = sec2 x

d
dx
cot x = − csc2 x
Key vocab, techniques
vocab: none
techniques: Apply the derivatives of the trigonometric functions in any
type of problem dealing with differentiation we’ve seen thus far.
Problems
1. Find a point P on the graph of f (x) = tan x such that the tangent line
to y = f (x) at P is parallel to the line 4x − 3y = 2.
1
Section 2.5 / The chain rule
Overview
We’ve seen how to differentiate some basic types of functions. One thing
we don’t know how to handle
yet is compositions. For example, what is the

2
derivative of f (x) = sin x + 1? It would be extremely difficult to use the
limit definition of a derivative to solve this problem. We learn the chain rule,
which teaches us how to differentiate compositions.
Details
discuss: A prerequisite skill: recognizing a composition of functions, and
breaking the composition into its components.
discuss: Substitution in differentiation using Leibniz notation.
th: (the chain rule) If h(x) = f (g(x)), then h0 (x) = f 0 (g(x))g 0 (x). In
Leibniz notation, this looks like
d(f (g(x))) d(g(x))
d(h(x))
=
·
dx
d(g(x))
dx
ex: Some examples of differentiating compositions of functions.
Key vocab, techniques
vocab: chain rule
techniques: Use the chain rule to differentiate compositions of functions.
Given a table of values for two functions f and g and their derivatives, use the
chain rule to answer questions about the values of the derivatives of f (g(x))
and g(f (x)).
Problems
1.
Let h(x) = f (g(x)). Use the following table to order from least to
greatest h0 (2), h0 (3), h0 (5), h0 (7), h0 (11).
1
x f (x)
2
3
3
5
5
7
7
11
11
2
f 0 (x)
5
7
11
2
3
g(x) g 0 (x)
7
11
11
2
2
3
3
5
5
7
2.
Let h(x) = f (g(x)). Use the following table to order from least to
greatest h0 (2), h0 (3), h0 (5), h0 (6), h0 (0).
x f (x)
6
5
3
6
0
3
2
0
5
2
f 0 (x) g(x) g 0 (x)
3
2
0
5
0
2
2
5
6
3
5
6
6
3
0
3.
Let h(x) = f (g(x)). Use the following table to order from least to
greatest f (0), g(1), h0 (2), h0 (3), h0 (4).
f 0 (x) g(x) g 0 (x)
2
3
4
3
4
0
4
0
1
0
1
2
1
2
0
p
4. Find the derivative of f (x) = tan( x2 − sin2 x).


5. Find the derivative of f (x) = 4 sec tan sin(x2 + 1) .
x f (x)
0
1
1
2
2
3
3
4
4
0

6. Compute the derivative of f (x) = sin x2 + x − 2.
7. Compute the derivative of f (x) =
8. Find the derivative:
p
3
sec3 (x − 1) + x3 .
d
sin3 x
dx
9. Find the derivative:
d
dx




2
cos sin cos x − x

2
10. Find the derivative:
d
dx
r
x+
q

x+ x
q
11. Find the derivative of f (x) = 3 sec
12. Find the derivative of f (x) =

3
x2 −1
2+cos x

.
1
.
x3 +sin x
13. Find the derivative of f (x) = sin4 (x4 ).
14. Find the derivative of f (x) = sin4 (sin x4 ).


15. Find the derivative of f (x) = 4 sin tan sec(x2 + 1) .
16. Let f (x) =
2
.
sin(3x)+2
(a) Find the smallest positive number c such that f 0 (c) = 0.
(b) Using your answer from part (a), find the equation of the line tangent
to the graph y = f (x) at the point (c, f (c)).
3
Section 2.6 / Implicit differentiation
Overview
Consider the equation of the circle x2 + y 2 = 1. We can view y as a function
of x, so it makes sense to talk about the derivative y 0 . However, if we want
to find the derivative, using the techniques available up til now we would
need to solve for y first. In some equations it is inconvenient or impossible
to solve for y. That won’t stop us from finding y 0 as long as we understand
implicit differentiation.
Details
discuss: Suppose y is a function of x. The chain rule applies to functions
of y like so:
d(f (y)) d(y)
d
f (y) =
·
.
dx
d(y)
dx
Exploitation of this fact is known as “implicit differentiation.”
rk: Often, when using implicit differentiation, the resulting formula for y 0
will involve both x and y. If you wish, you can solve for y in the original
equation, and substitute it in.
ex: Find y 0 : x2 + y 2 = 25.
rk: Sometimes it’s not possible to solve for y. We can still find y 0 .
ex: x3 + y 3 = 6xy.

ex: 2 x + y = 3
ex: If f (x) + x2 (f (x))3 = 10 and f (1) = 2, find f 0 (1).
ex: Find the equation of the tangent line to the hyperbola
the point (x0 , y0 ).
Key vocab, techniques
vocab: implicit differentiation
1
x2
a2

y2
b2
= 1 at
techniques: Perform implicit differentiation in order to find derivatives of
functions without needing to solve for y. Apply this to all sorts of problems
involving finding derivatives (finding equations of tangent lines, finding where
a graph is increasing, etc).
Problems
1. Find the tangent line to the graph of x2 + y 2 = 5×4 at the point (1, 2).
2. Find the equation of the tangent line to the graph y = (x5 −31)32 at x = 2.

3. Find the tangent line to the graph of x + (xy)3 = 12 at 4, 21 .
4. Find the equation of the tangent line to x cos(xy) =
π

2
at (π, 41 ).
5. Find the derivative of y 2 (y 2 − 4) = x2 (x2 − 5) at the point (0, −2).
6. Find the derivative of 2(x2 + y 2 )2 = 25(x2 − y 2 ) at (3, 1).

7. Find the derivative of x2/3 + y 2/3 = 4 at (−3 3, 1).

8. Find the derivative of x2 + y 2 = (2×2 + 2y 2 − x)2 at 0, 21 .
2

Purchase answer to see full
attachment

error: Content is protected !!