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Derivative. x. b - a. 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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x b - a 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
This average rate of change the diff quotient of f over the interval [a, b].
4 - 2 (a+h) - a 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 +h2+1
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
Before Zooming In After Zooming In Notice how the curve appears to "flatten" as we zoom 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. Sol:
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.
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. Therefore, the derivative is a function of x, and that is why we write it as f'(x)
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 by:
Negative Exponents Since the power rule works for negative exponents, 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."
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 0 h 0 h 0 h hg(x)g(x+h) hg(x)g(x+h) g(x)g(x+h)
d (f(x)/g(x)) = g(x).f'(x) - f(x).g'(x) dx g(x).g(x)
Ex: Ex: Ex:
Limits and Continuity: Numerical Approach Estimating Limits Numerically "What happens to f(x) as x approaches 2?" Calculatingthe limit of f(x) as x approaches 2,
Ex:lim f(x) = ? x 3 Ex: lim g(x) = lim g(x) = lim g(x) = - + x x x -5 -5 -5
Estimate x -2 f(x) Limits and Continuity: Graphical Approach
lim f(x) x -2 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
lim f(x) x a+ 3. Move the point along the graph toward x = a from the right . The value the y-coordinate approaches (if any) is
lim f(x)=2 x -2+ The y-coordinate is approaching 2 as x approaches -2 from the right. Therefore,
lim f(x)= x a- 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
lim f(x)=2 x -2- The y-coordinate is again approaching 2 as x approaches -2 from the left.
lim f(x)=2 x a 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
X2-3x 2x+3 X2-3x 2x+3 x 2 = -0.1818181…=-2/11 f(2)= -0.1818181…=-2/11 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)=
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. x a
x+2 x -2 Let us evaluate lim 3x2+x-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)? -11
d dx (x2/x5)=2x/5x4 is: The statement 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 ofax2+bx+c=0.
We are going to solve for x. ax2+bx+c=0 Divided through by a. x2+ b/a x+ c/a=0 Subtractedc/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