Transcript Lesson 6-2

Lesson 3-R
Review of Derivatives
Objectives
• Find derivatives of functions
• Use derivatives as rates of change
• Use derivatives to find related rates
• Use derivatives to approximate change in
dependent variables (y)
Vocabulary
• None new
Basic Differentiation Rules
d
---- (c) = 0
dx
Constant
d
---- (xⁿ) = nxn-1
dx
Power Rule
d
d
---- [cf(x)] = c ---- f(x)
dx
dx
Constant Multiple Rule
d
---- (ex) = ex
dx
Natural Exponent
d
1
---- (ln x) = ----dx
x
Natural Logarithms
Other Differentiation Rules
Constant to Variable Exponent Rule
d
----- [ax] = ax ln a
dx
This is a simple example of logarithmic differentiation
that we will examine in a later problem.
Sum and Difference Rules
d
d
d
---- [f(x) +/- g(x)] = ---- f(x) +/- ---- g(x)
dx
dx
dx
In words: the derivative can be applied across an
addition or subtraction. This is not true for a
multiplication or a division as the next two rules
demonstrate.
Product Differentiation Rule
d
d
d
---- [f(x) • g(x)] = f(x) • ---- g(x) + g(x) • ---- f(x)
dx
dx
dx
In words: the derivative of a product of two functions is
the first function times the derivative of the second
function plus the second function times the derivative of
the first function.
Quotient Differentiation Rule
d
d
g(x) ----- [f(x)] – f(x) -----[g(x)]
d
f(x)
dx
dx
---- [--------] = -----------------------------------------------dx g(x)
[g(x)]²
In words: the derivative of a quotient of two functions is
the denominator times the derivative of the numerator
minus the numerator times the derivative of the
denominator, all divided by the square of the denominator.
Trigonometric Functions
Differentiation Rules
d
---- (sin x) = cos x
dx
d
---- (cos x) = –sin x
dx
d
---- (tan x) = sec² x
dx
d
---- (cot x) = –csc² x
dx
d
---- (sec x) = sec x • tan x
dx
d
---- (csc x) = –csc x • cot x
dx
Hint: The derivative of trig functions (the “co-functions”) that begin
with a “c” are negative.
Derivatives of Inverse
Trigonometric Functions
d
1
-1
---- (sin x) = -----------dx
√1 - x²
d
-1
-1
---- (cos x) = ----------dx
√1 - x²
d
1
-1
---- (tan x) = ------------dx
1 + x²
d
-1
-1
---- (cot x) = ------------dx
1 + x²
d
1
-1
---- (sec x) = -------------dx
x √ x² - 1
d
-1
-1
---- (csc x) = ------------dx
x √ x² - 1
Interesting Note:
If f is any one-to-one differentiable function, it can be proved that its inverse
function f-1 is also differentiable, except where its tangents are vertical.
Differentiation Chain Rule
What if I have something other than just x in one of the
previous formulas? If f and g are both differentiable and
F = f○g is the composite function defined by F(x) = f(g(x))
d
---- [F(x)] = f’(g(x)) • g’(x)
dx
or
dy
dy
du
---- = ---- • ---dx
du
dx
In words: the derivative of a composite function is equal to
the derivation of the outer function times the derivative of
the inner function.
The notation on the right is Leibniz notation and is often
referred to as u substitution. By letting u=g(x) we change
f(g(x)) to f(u). Then its derivative is chained by derivative of
y with respect to u multiplied by the derivative of u with
respect to x.
Example 1
Find the derivatives of the following:
1. f(t) = 7t³ – 4t + 12π
f’(t) = 21t² - 4
We used constant rule,
constant multiple rule,
and power rule.
2. f(x) = csc (x)
f’(x) = -csc (x) cot (x)
We used trig rule for csc.
Remember we never change
what is inside the trig func.
Example 2
Find the derivatives of the following:
3. f(x) = 12(3x + 1)4 + sin (5x²) + 7
f’(x) = 48(3x +1)³ (3) + cos (5x²) (10x)
We used chain rule with u=3x+1 in the first term and v=5x² in the
second term. This gives us 4∙12(u)³∙(u’) + sin(v)∙(v’).
4. y = ln(4x + 1) + e-6x
y’(x) = (4/(4x + 1)) + -6e-6x
We used chain rule with u=4x+1 in the first term and v=-6x in the
second term. This gives us (u’)/ u + (v’)ev. Note we do not change
the exponent when it has the variable in it.
Example 3
Find the derivatives of the following:
5. d(t) = cot-1(et)
-et
d’(t) = --------------1 + (et)²
We use chain rule with u=et using inverse trig derivative.
This gives us -(u’) / (1 + u²)
6. g(t) = (7t4 – 4t3) (6t2 + 1)²
g’(t) = (28t³ - 12t²)(6t² + 1)² + (7t4 – 4t3) (2)(6t2 + 1) (12t)
We used product rule and u=6t²+1 for the second term (applying
chain rule).
Example 4
Find the derivatives of the following:
7. y = (2e3x – cos(4x)) / (7x² – 9x³)
(7x² - 9x³)[2e3x(3) - -4sin(4x)] – (14x – 27x²[2e3x – cos(4x)]
y’(x) = -------------------------------------------------------------------------------(7x² - 9x³)²
We use quotient rule and need to use chain rule twice with
the derivative of the numerator.
8. f(x) = 0
f’(x) = 0
Implicit Differentiation
If a function (or a relation) can not be set into the for y =
f(x), then implicit differentiation using the chain rule to find
y’ (differentiating both sides with respect to x and solving
for y’) can be used to find the derivative.
Example: (a circle with radius 10, which is a relation and
not a function) x² + y² = 100
dx
dy
2x --- + 2y ---- = 0
dx
dx
dy
2x = -2y ---dx

dy
2x + 2y ---- = 0
dx
dy
-x
 ---- = ----dx
y
Logarithmic Differentiation
Steps in Logarithmic Differentiation:
1. Take natural log of both sides of an equation y = f(x)
2. Use to laws of logs to simplify
Laws of Logarithms:
loga (xy) = loga x + loga y
loga (x/y) = loga x - loga y
loga xr = r loga x
(where r is a real number)
3. Differentiate implicitly with respect to x
4. Solve the resulting equation for y’ (dy/dx)
5. Substitute back in what y was in step 1.
(product)
(quotient)
(exponent)
Example 5
Find the derivatives of the following:
11. xy² + y³ - 6x = 25
y² + x(2yy’) + 3y²y’ – 6 = 0
We need to implicitly differentiate using
the chain rule to get y’. Separate terms
involving y’ from the rest. Then divide
out to get y’ by itself
y’[2xy + 3y²] = 6 - y²
y’ = (6 - y²) / [2xy + 3y²]
12. y = 4x
ln y = ln 4x
ln y = x ln 4
y’/y = ln 4
We use logarithmic differentiation in this
problem. First we take the ln (natural log) of
both sides. Then we use the laws of logs to
simplify. Next take the derivative implicitly of
both sides. Substitute back in for y.
so y’ = y (ln 4)
and y’ = 4x (ln 4)
Related Rates
1. Let t = elapsed time and draw a diagram of the given situation
for t > 0. Be sure to draw the picture carefully and accurately
and include all of the variables in the problem. Assign
appropriate variables to the other quantities that vary with t and
label the diagram appropriately. Some dimensions in the
problem remain fixed as time passes. Label these as constants
in the diagram. Other information defines the point in time at
which you are to calculate the rate of change. Do not label
these dimensions as constants as they vary with time.
2. Identify what is given and what is wanted in terms of the
established variables.
3. Write a general equation relating the variables.
4. Differentiate this equation with respect to t.
5. Substitute the known quantities identified in step 2 into your
equation and find the solution to the problem.
Related Rates Example
A small balloon is released at a point 30 feet away from
an observer, who is on level ground. If the balloon goes
straight up at a rate of 9 feet per second, how fast is the
distance from the observer to the balloon increasing
when the balloon is 40 feet high?
Base Equation: Pythagorean Thrm
d² = g² + h²
looking for dd/dt
d
h = 40
2d (dd/dt) = 2g (dg/dt) + 2h (dh/dt)
2(50) (dd/dt) = 2(30)(0) + 2(40) (9)
(dd/dt) = 2(40)(9) / (2) (50) = 7.2 ft/sec
dh
---- = 9
dt
θθ

30

Linear Approximation and Differentials
Linear Approximation  L(x) = f(a) + f’(a)(x – a)
= f(a) + f’(a)(∆x)
Makes a line (L) using the slope of the
function at point a to estimate small
changes around a. It uses the
alternate form of the derivative:
L(x)
f(x)
L(b)
f(b)
∆y
dy
f(a)
∆y
f(x) – f(a)
f’(a) = lim ----- = lim -------------∆x→0
∆x x→a
x-a
∆y = f(b) – f(a)
dx = ∆x
a
b
dy = L(b) – f(a)
dy
----- = f’(a)
dx
dy = f’(a) dx
For small changes in x (∆x close to 0), differential dy approximates ∆y
Differential Example
Use differentials to approximate the increase in the
volume of a cube when its side increases from 2 inches
to 2.05 inches.
Base Equation: Volume of a cube
V = s³
ds = ∆s = 0.05
dV ≈ ∆v
dSA = 3s² ds
∆V ≈ dV = 3(2)² (0.05) = 0.6 cu in
Summary & Homework
• Summary:
– Know derivative rules
• Use u-substitution to help with complex functions
• Use implicit differentiation when you can’t solve for y =
– Derivatives are rates of change (slope)
– When taking derivatives with respect to time
remember the product rule with more than one
variable changing with respect to time
• Homework:
– Study for part 1 of Chapter 3 test on derivatives