2 Average Rate of Change of f over [a, b]: Difference Quotient The average rate of change of the...

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Average Rate of Change of f over [a, b]: Difference Quotient

The average rate of change of the function f over the interval [a, b] is

Average rate of change of f = f = f(b) - f(a)

=Slope of line through points P and Q in the figure

x b - a

This average rate of change the diff quotient of f over the interval [a, b].

Ex: Let f(x) = x3 + x. Then: Rate of change of f over [2, 4]= f(4) - f(2) = 68-10/2=29

Rate of change of f over [a, a+h]= f(a+h) - f(a) = 3a2 + 3ah

(a+h) - a

A Numerical Approach

In Indonesia, you monitor the value of the US Dollar on the foreign exchange

market very closely during a rather active five-day period. Suppose

R(t) = 7,500 + 500t - 100t2 rupiahs, The rupiah is the Indonesian currency, where t

is time in days. (t = 0 represents the value of the Dollar at noon on Monday.)

What was the value of the Dollar at noon on Tuesday?

According to the graph, when was the value of the Dollar rising most rapidly?

A formula for the average rate of change of the Dollar's value over the interval [1, 1+h] is given by

Use your answer to the last question to complete the following table.

Instantaneous Rate of Change of f(x)

at x = a: The Derivative

The instantaneous rate of change of f(x) at x = a is defined by taking the limit of the average rates of change of f over the intervals [a, a+h], as h approaches 0.

The instantaneous rate of change the derivative of f at x = a which we write as f'(a).

The Derivative as Slope: A Geometric Approach

Estimating the Slope by Zooming In

Notice how the curve appears to "flatten" as we zoom in;

Before Zooming In After Zooming In

Slope of the Secant Line and Slope of the Tangent Line

The slope of the secant line through

(x, f(x)) and (x+h, f(x+h)) is the same as the average rate of change of f over the interval [x, x+h], or the difference quotient:

The slope of the tangent line through (x, f(x)) is the same as the instantaneous rate of change of f at the point x, or the

derivative:

Ex: Let f(x) = 3x2 + 4x. Use a difference quotient with h = 0.0001 to estimate the slope of the tangent line to the graph of f at the point where x = 2.

The Derivative as a Function:

An Algebraic Approach

So far, all we have been doing is approximating the derivative of a function. Is there a way of computing it exactly?

Recall: The derivative of the function f at the point x is the slope of the tangent line through (x, f(x)), or the instantaneous rate of change of f at the point x.

The slope of the tangent, or derivative, depends on the position of the point P on the curve, and therefore on the choice of x.

Therefore, the derivative is a function of x, and that is why we write it as f'(x)

f'(1) = slope of the tangent at the point on the graph where x = 1.

f'(-4) = slope of the tangent at the point on the graph where x = -4.

Definition

The derivative f'(x) of the function f(x) is the slope of the tangent at the point (x, f(x)).

In words, the derivative is the limit of the difference quotient.

By the "difference quotient" we

mean the average rate of change of f over the interval [x, x+h]:

Definition

A derivative f'(x) of a function f depicts how the function f is changing at point x.

f must be continuous at point x in order for there to be a derivative at that point. A function which has a derivative is said to be differentiable.

The derivative is computed by using the concept of x. x is an arbitrary change or increment in the value of x.

Ex: Let f(x) = 3x2 + 4x. The difference quotient is given by:

Hint: Average rate of change of f over [x, x+h] =

Now take the limit as h 0.

Ex Continued : f'(x) = f'(1) =

Ex: Let f(x) = 1/x, f(x+h) is given by The difference quotient is given

Power Rule

Negative Exponents Since the power rule works for

negativeexponents, we have, for

Ex: If f(x) = x3, then f'(x) = 3x2. When we say "f'(x) = 3x2," "The derivative of x3 with respect

to x equals 3x2." “The derivative with respect to x"

by the symbol "d/dx."

Derivatives of Sums, Differences & Constant Multiples

The quotient f(x)/g(x)If f(x) and g(x) are diff. Then d (f(x)/g(x)) = lim f(x+h)/g(x+h) -(f(x)/g(x))

= lim f(x+h).g(x) - f(x).g(x+h)

= lim f(x+h).g(x) -f(x)g(x) +f(x)g(x) - f(x).g(x+h)

= lim g(x) . (f(x+h)-f(x))/h - f(x) . (g(x+h)-g(x))/h

dx h 0 h

h 0 hg(x)g(x+h)

h 0 g(x)g(x+h)

dxg(x).f'(x) - f(x).g'(x) d (f(x)/g(x)) =

g(x).g(x)

Limits and Continuity: Numerical Approach

Estimating Limits Numerically"What happens to f(x) as x approaches

Calculating the limit of f(x) as x approaches 2,

lim g(x) = -5x - lim g(x) =

-5x lim g(x) = -5x +

Ex: lim f(x) = ?x 3

Limits and Continuity: Graphical

Approach Estimate x -2 f(x)

To decide whether x a f(x) exists, and to find its value if it does.

1. Draw the graph of f(x) either by hand or using a graphing calculator.

2. Position your pencil point (or the graphing calculator "trace" cursor) on a point of the graph to the right of x = a. In the example illustrated, we are estimating

-2f(x)

3. Move the point along the graph toward x = a from the right . The value the y-coordinate approaches (if any) is lim

a+f(x)

The y-coordinate is approaching 2 as x approaches -2 from the right. Therefore, lim

-2+f(x)=2

Repeat Steps 2 and 3, but this time starting from a point on the graph to the left of x = a, and approach x = a along the graph from the left. The y-coordinate approaches (if any) is then

a-f(x)=

The y-coordinate is again approaching 2as x approaches -2 from the left.

-2-f(x)=2

5. If the left and right limits both exist and have the same value L, then

lim f(x) = exists and equals L.

The left and right limits both exist and equal 2, and so

-2f(x)=2

Limits and Continuity: Algebraic Approach

limx 2

2x+3=-0.1818181…=-2/11

notice that you can obtain the same answer bysimply substituting x = 2 in the given function:

f(x)=X2-3x

-0.1818181…=-2/11

Is that all there is to evaluating limits algebraically:

just substitute the number x is approaching in the given expression?

Ans: The function is continuous at the value of x in question.

Continuous Functions

The function f(x) is continuous at x=a if

lim f(x) exists and equals f(a).

The function f is said to be continuous on its domain if it is continuous at each point in its domain. If f is not continuous at a particular a, we say that f is discontinuous at a or that f has a discontinuous at a.

Let us evaluate lim 3x2+x-10x -2 x+2Ask yourself the following questions:

Is the function f(x) a closed form function?

Is the value x = a in the domain of f(x)?

The statement ddx

(x2/x5)=2x/5x4 is:

Wrong, because the correct answer is

(a) -3/x4

(b) 0/3x2 =0

(c) 1/3x2

(d) lnx3

Quadratic formula - derivation

For quadratic equations of the type

x2+ p x + q = 0

The derivation of the quadratic formula for the roots of ax2+bx+c=0.

We are going to solve for x. ax2+bx+c=0

Divided through by a. x2+ b/a x+ c/a=0Subtracted c/a on both sides. x2+ b/a x = -c/aComplete the square on the left. x2+ b/a x + (b/2a)2= -c/a + (b/2a)2

The left is square

(x+ b/2a)2 = -c/a + (b/2a)2

Common dominator is 4a2

(x+ b/2a)2 = -4ac+b2/4a2

Now take the square roots.(x+ b/2a)2 = b2-4ac

(x+ b/2a) = √ Subtract b/2a on both sides

X = - b/2a √

+ b2-4ac2a

The Product Rule

The derivative of a product is NOT the product of the derivatives.

In the form of u(x+h)v(x) -u(x+h)v(x) to the numerator

In the last step,

because u(x) is differentiable at x and therefore continuous.

The product u(x)v(x) as the area of a rectangle with width u(x) and height v(x). The change in area is d(uv), and is indicated is the figure below.

As x changes, the area changes from the area of the red rectangle, u(x)v(x), to the area of the largest rectangle, the sum of the read, green, blue and yellow rectangles. The change in area is the sum of the areas of the green, blue and yellow rectangles,

Product and Quotient Rule Product Rule

Quotient Rule

Ex: Find the derivative of

f(x) =(4x3-x4)(11x-√x).

Sol: First recognize that f(x) is a product of two factors: (4x3-x4) and (11x-√x)

Rewrite the function in exponent formf(X)= (4x3-x4)(11x-x0.5)

(4x3-x4)(11x-x0.5)

=(12x2-4x3) (11x-x0.5) +(4x3-x4) (11-0.5x-

In the limit of dx small, the area of the yellow rectangle is neglected. Algebraically,

Ex: For c is a constant,

Whether or not this is substantially easier than multiplying out the polynomial and differentiating directly is a matter of opinion; decide for yourself.

Ex: If f and g are differentiable functions such that f(2)=3, f’(2)=-1, g(2)=-5 and

g’(2)=2, then what is the value of (fg)’(2)?

Ex: With g(x)=(x3-1)(x3+1) what is g ’(x)?

EX: Find dy/dx where y(x)= (8x-1) (x2+4x+7)(x3-5)

Ex: If f, g and h are differentiable, use the product rule to show that

As a corollary, show that

Ex: ddx

(x2-x)(5+x-0.5)

Ex: ddx

(x2-1/x2)(5+x-0.5)

Ex: The derivative of f(x) = x2+3x+2 is?

Using the Calculation Thought Experiment (CTE)

Let us use the CTE to find the derivative of

f(x)=(3x+1) x2+4 x2+x

Ex: Find out the derivative of (3x-2 + 2/x)(x + 1)

Ex: Find out the derivative of f(x)=4x2+(x-1)(4x+1)

Suppose we want to find the derivative of y(x) = (x2+3x+1)2

We could hopefully multiply y(x) out and then take the derivative with little difficulty. But, what if,

y(x) = (x2+3x+1)50

Would you want to apply the same method to this problem?

Ex: Returning to the first y(x) above, if we let

The function y(x) is the composite of g with f.

Our goal is to find the derivative Based on our knowledge of the functions f and g. Now, we know that

Leading to the speculation that

This leads to the (possible) chain rule:

Ex: The function sin(2x) is the composite of the functions sin(u) and u=2x. Then,

Ex: sin2(4x) is a composite of three functions : u2, u=sin(v) and v=4x.

As a check, you may want to note that the above may be expressed as

Inverse Function: To find the derivative based on the knowledge or condition that

for some function f(t), or, in other words, that g(x) is the inverse of f(t) = x.

Recognizing that t and g(x) represent the same quantity, and remembering the Chain Rule,

This result becomes somewhat obvious;

We know from the Power Rule, with n=2, that

Equivalently

The above may be generalized; for nonzero n,

Rule name (if any)

The Sum rule

The Product rule

The Quotient rule

The Chain rule

The Power rule

Derivatives of Hyperbolic Functions:

Taking f(x)=x3 we get

U3= 3u2ddx

The Chain Rule

If u is a diff. function of x, and f is a diff. function of u, then:

[f(u)]= f’(u)ddx

Ex: ddx

[ ] 5 = X2-x-13x-1

[(x2 - 1)3(3x + 4)-

Derivatives of Logarithms:

Let u be a function of x,

Derivatives of Exponential Functions:Let u be a function of x,

Derivatives of Inverse Hyperbolic Functions:If u is a function of x;

Implicit Differentiation

Not all functions are given explicitly and are only implied by an equation.

Ex: xy = 1 is an equation given implicitly, explicitly it is y = 1/ x. Now to find dy/dx for xy = 1, simply solve for y and differentiate.

xy = 1y = 1 / x = x-1

dy/dx = -1 x-2= 1/x2

But, not all equations are easily solved for y, as in the equation 3x + y3 = y2

+ 4 This is where implicit differentiation is applied. Implicit differentiation is taking the derivative of both sides of the equation with respect to one of the variables.

Most commonly, used is the derivative of y with respect to x. or dy/dx.

Ex: 3x + y3 = y2 + 4, solve for dy/dx. 3x + y3 = y2 + 4

d/dx(3x + y3) = d/dx(y2 + 4)

3 + 3y2 dy/dx = 2y dy/dx

3 = 2y dy/dx - 3y2 dy/dx

3 = y ( 2 - 3y ) dy/dx

3 / y (2 - 3y ) = dy/dx

Ex: Find the slope of the curve x2 + y3 = 2x + y at ( 2,4) Sol: d/dx [x2 + y3] = d/dx [2x + y] 2x + 3y2 dy/dx = 2 + dy/dx 2x - 2 = (-3y2 + 1) dy/dx 2( x - 1) / (-3y2 + 1) = dy/dx = slope of curve substitute (2,4) into dy/dx to find the slope at that point. 2(2-1) / (-3 · 42 + 1) = 2 /-49 = -2/49 is the slope of the curve.

Derivatives of Higher Order

Derivatives of functions are also functions, therefore can be differentiated again.

Ex: f(x) = x5

f '(x) = 5x4

f ''(x) = 5·4x3 = 20x3

f '''(x) = 5·4·3x2 = 60x2

Maximum and Minimum Values of a FunctionSecond Derivative Test for FunctionsConcavity:

If the second derivative of a function f ( f ''(x) ) is positive (or negative) for all

x on (a,b) then the graph of f is concave

upward (or downward) on (a,b).

Second Derivative Test for Max. and Min. Points.If point A(a, f(a)) is on the graph of

function f such that f '(a) = 0 and

f ''(a) < 0 , then point A is a relative

maximum; if f '(a) = 0 and f ''(a) > 0 , then

point A is a relative minimum.

Maximum and Minimum Values of a Function

Increasing and Decreasing Functions:A function f is said to be increasing when f '(x) > 0 for every x on (a,b) and decreasing when f '(x) < 0 for every x on (a,b).Absolute Max. and Min. of a function:The absolute maximum (or minimum) of a function is a point (a, f(a)) when f(a) f(x) ( or f(a) < f(x) ) for any value of x in the domain of f.

Ex: Find the open intervals on which the function f(x) = x3 - 3x2 is increasing or decreasing.

Sol: f '(x) = 3x2 - 6x

let f '(x) = 3x2 - 6x = 0

x = 0 or 2 Critical numbers

Because there are no x-values for which f ' is undefined, it follows that x = 0 and x = 2 are the only critical numbers. So, the intervals that need to be tested are (- , 0), (0, 2), and (2, ).

Critical Point:

A critical point, (x,f(x)) , of a function f is if f(x) is defined and f '(x) is either zero or undefined. The x-coordinate of the critical point is called a critical value or a critical number.

2 Average Rate of Change of f over [a, b]: Difference Quotient The average rate of change of the...

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