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 Constantdx

d---- (xⁿ) = nxn-1 Power Ruledx

d d---- [cf(x)] = c ---- f(x) Constant Multiple Ruledx dx

d---- (ex) = ex Natural Exponentdx

d 1---- (ln x) = ----- Natural Logarithmsdx x

Other Differentiation RulesConstant to Variable Exponent Rule

d ----- [ax] = ax ln adx

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 d---- (sin x) = cos x ---- (cos x) = –sin xdx dx

d d---- (tan x) = sec² x ---- (cot x) = –csc² xdx dx

d d---- (sec x) = sec x • tan x ---- (csc x) = –csc x • cot xdx dx

Hint: The derivative of trig functions (the “co-functions”) that begin with a “c” are negative.

Derivatives of Inverse Trigonometric Functions

d 1 d -1---- (sin-1 x) = ------------ ---- (cos-1 x) = -----------dx √1 - x² dx √1 - x²

d 1 d -1---- (tan-1 x) = ------------- ---- (cot-1 x) = -------------dx 1 + x² dx 1 + x²

d 1 d -1---- (sec-1 x) = -------------- ---- (csc-1 x) = -------------dx x √ x² - 1 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 RuleWhat 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 dy dy du ---- [F(x)] = f’(g(x)) • g’(x) or ---- = ---- • ---- dx 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

1. f(t) = 7t³ – 4t + 12π

2. f(x) = csc (x)

Find the derivatives of the following:

f’(t) = 21t² - 4

f’(x) = -csc (x) cot (x)

We used constant rule,constant multiple rule, and power rule.

We used trig rule for csc.Remember we never changewhat is inside the trig func.

Example 2

3. f(x) = 12(3x + 1)4 + sin (5x²) + 7

4. y = ln(4x + 1) + e-6x

Find the derivatives of the following:

f’(x) = 48(3x +1)³ (3) + cos (5x²) (10x)

y’(x) = (4/(4x + 1)) + -6e-6x

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’).

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

5. d(t) = cot-1(et)

6. g(t) = (7t4 – 4t3) (6t2 + 1)²

Find the derivatives of the following:

-et

d’(t) = --------------- 1 + (et)²

g’(t) = (28t³ - 12t²)(6t² + 1)² + (7t4 – 4t3) (2)(6t2 + 1) (12t)

We use chain rule with u=et using inverse trig derivative. This gives us -(u’) / (1 + u²)

We used product rule and u=6t²+1 for the second term (applying chain rule).

Example 4

7. y = (2e3x – cos(4x)) / (7x² – 9x³)

8. f(x) = 0

Find the derivatives of the following:

(7x² - 9x³)[2e3x(3) - -4sin(4x)] – (14x – 27x²[2e3x – cos(4x)]y’(x) = -------------------------------------------------------------------------------- (7x² - 9x³)²

f’(x) = 0

We use quotient rule and need to use chain rule twice with the derivative of the numerator.

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

dy dy -x2x = -2y ---- ---- = ----- dx dx y

dx dy dy 2x --- + 2y ---- = 0 2x + 2y ---- = 0 dx dx dx

Logarithmic DifferentiationSteps in Logarithmic Differentiation:

1. Take natural log of both sides of an equation y = f(x)

2. Use to laws of logs to simplifyLaws of Logarithms:

loga (xy) = loga x + loga y (product)

loga (x/y) = loga x - loga y (quotient)

loga xr = r loga x (where r is a real number) (exponent)

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.

Example 5

11. xy² + y³ - 6x = 25

12. y = 4x

Find the derivatives of the following:

y² + x(2yy’) + 3y²y’ – 6 = 0

y’[2xy + 3y²] = 6 - y²

y’ = (6 - y²) / [2xy + 3y²]

ln y = ln 4x

ln y = x ln 4

y’/y = ln 4 so y’ = y (ln 4) and y’ = 4x (ln 4)

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

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.

Related Rates1. 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

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

d

30

θ

h = 40

dh ---- = 9dtθ

Linear Approximation and Differentials

Linear Approximation L(x) = f(a) + f’(a)(x – a) = f(a) + f’(a)(∆x)

a

f(a)

L(b)

b

dy----- = f’(a) dx

dy = f’(a) dx

∆y f(x) – f(a)f’(a) = lim ----- = lim -------------- ∆x x - a∆x→0 x→a

L(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:

f(x)

f(b)

dx = ∆x

∆y

∆y = f(b) – f(a)

dy = L(b) – f(a)

dy

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

dSA = 3s² ds

∆V ≈ dV = 3(2)² (0.05) = 0.6 cu in

dV ≈ ∆v

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