Derivatie class 12

Sadiq Hussain professor of mathematics à Manjanbazam Cadet College Tarbela Cantt.
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Derivatie class 12

• 4. Learning Objectives;  Students will be able to learn about;  What is Icrement.  What is meant by change in x i.e.(∆x).  What is meant by chane in y i.e .(∆y) .  What is average rate of change.  What is incremental ratio.  What is limit of function .  What is derivative of function.
• 5. What is Increment  The change in the values of ‘y’corresponding to small changes in the values of ‘x’ where ‘y’ &’x’ are related to each other by the equation; y=f(x). Explanation; let y=x^3.when x=2 we have y=8. if x changes from 2 to 2.1.Then increment in x=2.1-2=0.1. The corresponding value of y will be (2.1)^3.=9.261.
• 6. Contd..  Now if x changes from 2 to 0.5.Then increment in x=0.5-2=-1.5 Corresponding value of y at x=0.5 will be=0.125 Increment in y in this caseis 0.125-8=-7.875. Hence it is concluded now a small change in the value of x as it increases or decreases from one value of X=x1 to another value of X=x2. in its change is denoted by ∆x. (read as’Deltax’). And is called increment in x. Here ∆x= x2-x1 and we write x2=x1 + ∆x . Note. ∆x is not product of ∆ & x but merely a symbol for a small change in x.
• 7. The limit of a function  A quantity L is the limit of a function f (x) as x approaches a if, as the input values of x approach a (but do not equal a), th corresponding output values of f (x) get closer to L. Note that the value of the limit is not affected by the output value of f (x) at a. Both a and L must be real numbers. We write it as lim f (x) = L x → a
• 8. Incremental Ratio  Consider the function y=f(x).whose domain isreal no IR. As x changes to x+∆x,there will be a corresponding change in y of amount,say ∆y. Therefore,y will change from y to y+ ∆y. y+ ∆y=f(x+∆x). ∆y.= f(x+∆x)-y (by subtracting y b.s.). or. ∆y.= f(x+∆x)-f(x). Eq(1).
• 9. Icremental ratio contd.  Now dividing both side equation (1). By ∆x we get.  The above ratio of increment in y corresponding to the increment in x is called the incremental ratio or the difference quotient. Now we learn the pattern of change in the ratio corresponding to change in x.
• 10. Explanation  Let y=x^2.  Let us take initially x=3.  we take ∆x=4,3,2,1,0.1,0.01,0.001,……..etc. The following table gives the corresponding of ∆y and that of . View table on next slide in detail.
• 12. Observation  From above table we observe that ∆x decreases and approaches 0,∆y also dcreases and approaches 0.But the icremental ratio does not approach 0.instead it approaches a definite value 6. Hence we can conclude that lim = 6 x → a
• 13. Derivative of f(x).  The limit to the ratio as ∆x tends to 0.is called the derivative or differential coeffcient of y w.r.to x. Definition; Let y=f(x) be a continuous function of ‘ x’. Limit ∆x → 0 i.e. , if it exists, is called the differential co-efficient of f(x) w.r.t.’x’and may be denoted by dy/dx, d/dx(y), Dxy, d/dx(f(x)).
• 14. How to…  Given a function f, find the derivative by applying the definition of the derivative. 1. Calculate f (a + h). 2. Calculate f (a). 3. Substitute and simplify f (a + h) − f (a) h 4. Evaluate the limit if it exists: f′(a) = lim f (a + h) − f (a) h →0 h
• 15. Procedure of pattern of Derivative
• 17. Example-2  Finding the Derivative of a Polynomial Function Find the derivative of the function  f (x) = x2 − 3x + 5 at x = a.  Solution We have  f′(a) = lim f (a + h) − f (a) [Definition of a derivativ] h →0 h  Substitute f (a + h) = (a + h)2 − 3(a + h) + 5 and  f (a) = a2 − 3a + 5 in the formula
• 19. Try It #2  Find the derivative of the function f (x) = 3x2 + 7x at x = a. By formula.
• 20. Finding Derivatives of Rational Functions  To find the derivative of a rational function, we will sometimes simplify the expression using algebraic techniques we have already learned.
• 21. Example 3;  Finding the Derivative of a Rational Function Find the derivative of the function  f (x) = 3 + x ,at x = a. Sol. 2 − x
• 23. Try It #3  Find the derivative of the function  f (x) = 10x + 11 ,at x=a 5x + 4
• 24. Finding Derivatives of Functions with Roots  To find derivatives of functions with roots, we use the methods we have learned to find limits of functions with roots,including multiplying by a conjugate.
• 25. Example 4;  Finding the Derivative of a Function with a Root Find the derivative of the function f (x) = 4√ x at x = 36. Solution; We have
• 27. Try It #4  Find the derivative of the function f (x) = 9√ x at x = 9. by following the same above procedure.
• 28. Finding the Average Rate of Change of a Function  The functions describing the examples above involve a change over time. Change divided by time is one example of a rate. The rates of change in the previous examples are each different. In other words, some changed faster than others. If we were to graph the functions, we could compare the rates by determining the slopes of the graphs.
• 29. Concept of a tangent to a curve  A tangent line to a curve is a line that intersects the curve at only a single point but does not cross it there. (The tangent line may intersect the curve at another point away from the point of interest.) If we zoom in on a curve at that point, the curve appears linear, and the slope of the curve at that point is close to the slope of the tangent line at that point.
• 30. Contd.  Figure represents the function f (x) = x3 − 4x. We can see the slope at various points along the curve. • slope at x = −2 is 8 • slope at x = −1 is –1 • slope at x = 2 is 8
• 31. Explanation with Graph  Let’s imagine a point on the curve of function f at x = a as shown in Figure . The coordinates of the point are (a, f (a)). Connect this point with a second point on the curve a little to the right of x = a, with an x-value increased by some small real number h. The coordinates of this second point are ((a + h, f (a + h)) for some positive-value h.
• 32. Graph Connecting point a with a point just beyond allows us to measure a slope close to that of a tangent line at x = a
• 33. Contd..  We can calculate the slope of the line connecting the two points (a, f (a)) and (a + h, f (a + h)), called a secant line, by applying the slope formula, slope = change iny change in x We use the notation msec to represent the slope of the secant line connecting two points.
• 35. Example 1;  Finding the Average Rate of Change Find the average rate of change connecting the points (2, −6) and (−1, 5). Solution; We know the average rate of change connecting two points may be given by  If one point is (2, − 6), or (2, f (2)), then f (2) = −6. The value h is the displacement from 2 to −1, which equals −1 − 2 = −3. For the other point, f (a + h) is the y-coordinate at a + h, which is 2 + (−3) or −1, so f (a + h) = f (−1) = 5.
• 37. Try It #1  Find the average rate of change connecting the points (−5, 1.5) and ( − 2.5, 9).?.
• 38. Understanding the Instantaneous Rate of Change  Now that we can find the average rate of change, suppose we make h in Figure 2 smaller and smaller. Then a + h will approach a as h gets smaller, getting closer and closer to 0. Likewise, the second point (a + h, f (a + h)) will approach the first point, (a, f (a)). As a consequence, the connecting line between the two points, called the secant line, will get closer and closer to being a tangent to the function at x = a, and the slope of the secant line will get closer and closer to the slope of the tangent at x = a. See
• 39. Graph The connecting line between two points moves closer to being a tangent line at x = a.
• 40. Explanation of IROC  Because we are looking for the slope of the tangent at x = a, we can think of the measure of the slope of the curve of a function f at a given point as the rate of change at a particular instant. We call this slope the instantaneous rate of change, or the derivative of the function at x = a. Both can be found by finding the limit of the slope of a line connecting the point at x = a with a second point infinitesimally close along the curve. For a function f both the instantaneous rate of change of the function and the derivative of the function at x = a are written as f′(a), and we candefine them as a two-sided limit that has the same value whether approached from the left or the right.  The expression by which the limit is found is known as the difference quotient.
• 41. Finding Instantaneous Rates of Change  Many applications of the derivative involve determining the rate of change at a given instant of a function with the independent variable time—which is why the term instantaneous is used. Consider the height of a ball tossed upward with an initial velocity of 64 feet per second, given by,  s(t) = −16t2 + 64t + 6, where t is measured in seconds and s(t)is measured in feet. We know the path is that of a parabola. The derivative will tell us how the height is changing atany given point in time. The height of the ball is shown in Figure as a function of time. In physics, we call this the
• 43. Example ;  Finding the Instantaneous Rate of Change Using the function above, s(t) = −16t2 + 64t + 6, what is the instantaneous velocity of the ball at 1 second and 3 seconds into its flight? Solution ;The velocity at t = 1 and t = 3 is the instantaneous rate of change of distance per time, or velocity. Notice that the initial height is 6 feet. To find the instantaneous velocity, we find the
• 45. Conclusion;  s′(t) = − 32t + 64 Evaluate the limit by letting h = 0. For any value of t, s′(t) tells us the velocity at that value of t. Evaluate t = 1 and t = 3. s′(1) = −32(1) + 64 = 32 s′(3) = −32(3) + 64 = −32 The velocity of the ball after 1 second is 32 feet per second, as it is on the way up. The velocity of the ball after 3 seconds is −32 feet per second, as it is on the way down.
• 46. Try It #5  The position of the ball is given by  s(t) = −16t2 + 64t + 6. What is its velocity after 2 seconds into flight?
• 47. Using Graphs to Find Instantaneous Rates of Change  We can estimate an instantaneous rate of change at x = a by observing the slope of the curve of the function f (x) at x = a. We do this by drawing a line tangent to the function at x = a and finding its slope.
• 48. How To…  Given a graph of a function f (x), find the instantaneous rate of change of the function at x = a. 1. Locate x = a on the graph of the function f (x). 2. Draw a tangent line, a line that goes through x = a at a and at no other point in that section of the curve. Extendthe line far enough to calculate its slope as  change in y
• 49. Example;  Estimating the Derivative at a Point on the Graph of a Function From the graph of the function y = f (x) presented in Figure , estimate each of the following: f (0); f (2); f′(0); f′(2)
• 50. Solution ;  To find the functional value, f (a), find the y- coordinate at x = a. To find the derivative at x = a, f′(a), draw a tangent line at x = a, and estimate the slope of that tangent line. See Figure.
• 51.  a. f (0) is the y-coordinate at x = 0. The point has coordinates (0, 1), thus f (0) = 1. b. f (2) is the y-coordinate at x = 2. The point has coordinates (2, 1), thus f (2) = 1. c. f′(0) is found by estimating the slope of the tangent line to the curve at x = 0. The tangent line to the curve at x = 0 appears horizontal. Horizontal lines have a slope of 0, thus f′(0) = 0. d. f′(2) is found by estimating the slope of the tangent line to the curve at x = 2. Observe the path of the tangent line to the curve at x = 2. As the x value moves one unit to the right,
• 52. Try It #6  Using the graph of the function f (x) = x3 − 3x shown in Figure , estimate: f (1), f′(1), f (0), and f′(0).
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