Derivative

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Derivative
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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

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• Ex Let f(x) x3 x. Then
• Rate of change of f over
• 2, 4 f(4) - f(2) 68-10/229
• Rate of change of f over
• a, ah f(ah) - f(a) 3a2 3ah h21 

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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.)

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• 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?

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• A formula for the average rate of change of
  the Dollar's value over the interval 1, 1h is
  given by
• Use your answer to the last question to
  complete the following table.

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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, ah,
  as h approaches 0.

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

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The Derivative as Slope A Geometric Approach

• Estimating the Slope by Zooming In

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

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• Slope of the Secant Line and Slope of
  the Tangent Line
• The slope of the secant line through
• (x, f(x)) and (xh, f(xh)) is the same as the
  average rate of change of f over the interval x,
  xh, or the difference quotient

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• 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

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• 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.
• Sol

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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?

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• 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.

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• 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.

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• 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.
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Definition

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

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• 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,
  xh

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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.

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

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• Ex Let f(x) 3x2 4x. The difference quotient
  is given by
• Hint Average rate of change of f over x, xh
  
• Now take the limit as h 0.
• Ex Continued f'(x)
• f'(1)

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• Ex Let f(x) 1/x,
• f(xh) is given by
• The difference quotient is given by

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Power Rule
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• Negative Exponents
• Since the power rule works for negative
• exponents, we have, for

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• 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."

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Derivatives of Sums, Differences Constant
Multiples
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The quotient f(x)/g(x) If f(x) and g(x) are diff.
Then d (f(x)/g(x)) lim f(xh)/g(xh)
-(f(x)/g(x)) lim
f(xh).g(x) - f(x).g(xh) lim f(xh).g(x)
-f(x)g(x) f(x)g(x) - f(x).g(xh) lim g(x) .
(f(xh)-f(x))/h - f(x) . (g(xh)-g(x))/h
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• Ex

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Limits and Continuity Numerical Approach

• Estimating Limits Numerically
• "What happens to f(x) as x approaches 2?"

Calculating the limit of f(x) as x approaches 2,
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Limits and Continuity Graphical Approach
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lim

• To decide whether x    a f(x) exists, and to
  find its value if it does.
• Draw the graph of f(x) either by hand or using a
  graphing calculator.
• 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

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3. Move the point along the graph toward x a
from the right . The value the y-coordinate
approaches (if any) is
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• The y-coordinate is approaching 2 as x
  approaches -2 from the right. Therefore,

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• 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

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

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• 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

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Limits and Continuity Algebraic Approach

• lim

notice that you can obtain the same answer
by simply substituting x 2 in the given
function
f(x)
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• Is that all there is to evaluating limits
  algebraically
• just substitute the number x is approaching in
  the given expression?

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

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Continuous Functions The function f(x) is
continuous at xa 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.
x a
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• Let us evaluate
• lim 3x2x-10

Ask yourself the following questions Is the
function f(x) a closed form function? Is the
value x a in the domain of f(x)?
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The statement

• Wrong, because the correct answer is
• (a) -3/x4
• (b) 0/3x2 0
• (c) 1/3x2
• (d) lnx3

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• Quadratic formula - derivation
• For quadratic equations of the type
• x2 p x q 0
• The derivation of the quadratic formula for the
  roots of ax2bxc0.

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• We are going to solve for x. ax2bxc0
• Divided through by a.
• x2 b/a x c/a0
• Subtracted c/a on both sides.
• x2 b/a x -c/a
• Complete 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

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• Common dominator is 4a2
• (x b/2a)2 -4acb2/4a2
• Now take the square roots.
• (x b/2a)2 b2-4ac
• (x b/2a) v
• Subtract b/2a on both sides

X - b/2a v
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The Product Rule
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The derivative of a product is NOT the product of
the derivatives.
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In the form of u(xh)v(x) -u(xh)v(x) to the
numerator
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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.
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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,
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• Product and Quotient Rule
• Product Rule

Quotient Rule
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• Ex Find the derivative of
• f(x) (4x3-x4)(11x-vx).
• Sol First recognize that f(x) is a product of
  two factors (4x3-x4) and (11x-vx)
• Rewrite the function in exponent form
• f(X) (4x3-x4)(11x-x0.5)

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In the limit of dx small, the area of the yellow
rectangle is neglected. Algebraically,
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Ex For c is a constant,
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Ex
Whether or not this is substantially easier than
multiplying out the polynomial and
differentiating directly is a matter of opinion
decide for yourself.
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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)(x31) what is g (x)? EX Find dy/dx
where y(x) (8x-1) (x24x7)(x3-5)
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Ex If f, g and h are differentiable, use the
product rule to show that
As a corollary, show that
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• Ex

Ex
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• Ex The derivative of
• f(x) x23x2 is?

Using the Calculation Thought Experiment (CTE)
Let us use the CTE to find the derivative
of f(x)(3x1) x24
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• Ex Find out the derivative of
• (3x-2 2/x)(x 1)
• Ex Find out the derivative of  
• f(x)4x2(x-1)(4x1)  

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• Suppose we want to find the derivative of
• y(x) (x23x1)2
• We could hopefully multiply y(x) out and then
  take the derivative with little difficulty. But,
  what if,
• y(x) (x23x1)50
• Would you want to apply the same method to this
  problem?

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Ex Returning to the first y(x) above, if we let
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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
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This leads to the (possible) chain rule
Ex
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Ex The function sin(2x) is the composite of the
functions sin(u) and u2x. Then, Ex




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• Ex sin2(4x) is a composite of three functions
  u2, usin(v) and v4x.

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


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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,
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This result becomes somewhat obvious
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Ex
We know from the Power Rule, with n2, that
Equivalently
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The above may be generalized for nonzero n,
Then
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Derivatives of Hyperbolic Functions
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The Chain Rule If u is a diff. function of x, and
f is a diff. function of u, then
Taking f(x)x3 we get
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Ex
3x-1
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Derivatives of Logarithms Let u be a function of
x,                                              
      
Derivatives of Exponential Functions Let u be a
function of x,                                  
                     
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Derivatives of Inverse Hyperbolic Functions If u
is a function of x                             
         
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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-1dy/dx -1 x-2 1/x2
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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.
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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
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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.
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Derivatives of Higher Order
Derivatives of functions are also functions,
therefore can be differentiated again. Ex
f(x) x5 f '(x) 5x4 f
''(x) 54x3 20x3 f '''(x) 543x2
60x2
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Maximum and Minimum Values of a Function Second
Derivative Test for Functions Concavity 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).
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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) lt
0 , then point A is a relative maximum if f
'(a) 0 and f ''(a) gt 0 , then point A is a
relative minimum.
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Maximum and Minimum Values of a
Function Increasing and Decreasing FunctionsA
function f is said to be increasing when f '(x)
gt 0 for every x on (a,b) and decreasing when f
'(x) lt 0 for every x on (a,b).
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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, ).
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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.
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