Tuesday, Feb 12. These slides will cover the following. [cos(x)] = sin(x) 1 d. 2 higher-order derivatives. 3 tangent line problems

Transcription

1 Tuesday, Feb 12 These slides will cover the following. 1 d dx [cos(x)] = sin(x) 2 higher-order derivatives 3 tangent line problems 4 basic differential equations

2 Proof First we will go over the following derivative rule. Theorem d dx [cos(x)] = sin(x)

3 Proof First we will go over the following derivative rule. Theorem d dx [cos(x)] = sin(x) To see why this is true, we use the limit definition of the derivative. Let f (x) = cos(x) and we have f f (x + h) f (x) (x) = lim h 0 h cos(x + h) cos(x) = lim h 0 h

4 Proof First we will go over the following derivative rule. Theorem d dx [cos(x)] = sin(x) To see why this is true, we use the limit definition of the derivative. Let f (x) = cos(x) and we have f f (x + h) f (x) (x) = lim h 0 h cos(x + h) cos(x) = lim h 0 h Here we need to use the trigonometric identity cos(a + B) = cos(a) cos(b) sin(a) sin(b) with A = x and B = h. This means that cos(x + h) = cos(x) cos(h) sin(x) sin(h).

5 Proof So we get, f (x) = lim h 0 cos(x + h) cos(x) h cos(x) cos(h) sin(x) sin(h) cos(x) = lim h 0 h (cos(x) cos(h) cos(x)) sin(x) sin(h) = lim h 0 h ( ( ) cos(x) cos(h) cos(x) = lim h 0 h ( cos(x) cos(h) cos(x) = lim h 0 h ( cos(x)(cos(h) 1) = lim h 0 h ) lim h 0 ) lim h 0 sin(x) sin(x) sin(h) ) h ( ) sin(x) sin(h) h ( ) sin(h) h

6 Proof Since there are no h s in sin(x) or cos(x), we can pull them outside the limit like this, [ ] [ ] cos(h) 1 sin(h) cos(x) lim sin(x) lim h 0 h h 0 h sin(x) Using our special trigonometric limits lim x 0 x = 0, we get lim x 0 cos(x) 1 x [ cos(x) lim h 0 ] [ cos(h) 1 sin(x) lim h h 0 and hence f (x) = sin(x). = 1 and ] sin(h) = cos(x)[0] sin(x)[1] h = sin(x)

7 Higher-order derivatives First consider f (x) = x 5 5x 2 + sin(x). Now differentiate and we get f (x) = 5x 4 10x + cos(x) which we call the first derivative of f (x).

8 Higher-order derivatives First consider f (x) = x 5 5x 2 + sin(x). Now differentiate and we get f (x) = 5x 4 10x + cos(x) which we call the first derivative of f (x). If we differentiate f (x), then we get f (x) = 20x 3 10 sin(x) which we call the second derivative of f (x).

9 Higher-order derivatives First consider f (x) = x 5 5x 2 + sin(x). Now differentiate and we get f (x) = 5x 4 10x + cos(x) which we call the first derivative of f (x). If we differentiate f (x), then we get f (x) = 20x 3 10 sin(x) which we call the second derivative of f (x). And if we differentiate this again we get f (x) = 60x 2 cos(x) which we call the third derivative of f (x).

10 Higher-order derivatives First consider f (x) = x 5 5x 2 + sin(x). Now differentiate and we get f (x) = 5x 4 10x + cos(x) which we call the first derivative of f (x). If we differentiate f (x), then we get f (x) = 20x 3 10 sin(x) which we call the second derivative of f (x). And if we differentiate this again we get f (x) = 60x 2 cos(x) which we call the third derivative of f (x). Second, third, fourth, etc. derivatives are called higher-order derivatives.

11 Higher-order derivatives The following is a chart of the notation we use for higher order derivatives. function y y f (x) f (x) first derivative y dy dx f d (x) dx [f (x)] second derivative y d2 y f d (x) 2 [f (x)] dx 2 dx 2 third derivative y d3 y dx 3 f (x) d 3 dx 3 [f (x)] fourth derivative y 4 d4 y dx 4 f 4 (x) d 4 dx 4 [f (x)] fifth derivative y 5 d5 y dx 5 f 5 (x) d 5 dx 5 [f (x)].....

12 Tangent line problems You will find lots of practice problems and homework problems that simply ask you to differentiate. The following examples are to illustrate some of the types of tangent line problems that you may come across.

13 Tangent line problems First, let s take a look at a straightforward tangent line problem. Find an equation of the tangent line to f (x) = x + 3 at x = 4.

14 Tangent line problems First, let s take a look at a straightforward tangent line problem. Find an equation of the tangent line to f (x) = x + 3 at x = 4. To write the equation of the tangent line we will need a point and a slope.

15 Tangent line problems First, let s take a look at a straightforward tangent line problem. Find an equation of the tangent line to f (x) = x + 3 at x = 4. To write the equation of the tangent line we will need a point and a slope. point: At x = 1, f (4) = = 5, and so the point is (4, 5)

16 Tangent line problems First, let s take a look at a straightforward tangent line problem. Find an equation of the tangent line to f (x) = x + 3 at x = 4. To write the equation of the tangent line we will need a point and a slope. point: At x = 1, f (4) = = 5, and so the point is (4, 5) slope: We differentiate to get f (x) = 1 2 x 1/2, and then m = f (4) = 1 ( 1 )( 1 ) ( 1 )( 1 2 (4) 1/2 = 4 = = 2 2 2) 1 4

17 Tangent line problems First, let s take a look at a straightforward tangent line problem. Find an equation of the tangent line to f (x) = x + 3 at x = 4. To write the equation of the tangent line we will need a point and a slope. point: At x = 1, f (4) = = 5, and so the point is (4, 5) slope: We differentiate to get f (x) = 1 2 x 1/2, and then m = f (4) = 1 ( 1 )( 1 ) ( 1 )( 1 2 (4) 1/2 = 4 = = 2 2 2) 1 4 Therefore the equation of the tangent line is y 5 = 1 (x 4). 4

18 Tangent line problems Find an equation of the tangent line to f (x) = 3x 2 2 that is perpendicular to y = 2x + 5.

19 Tangent line problems Find an equation of the tangent line to f (x) = 3x 2 2 that is perpendicular to y = 2x + 5. Again, we need a point and a slope. Since we want a tangent line that is perpendicular to y = 2x + 5, which has a slope of 2. Since perpendicular lines have negative reciprocal slopes, we want a slope of 1 2. So we have our slope, in this problem we have to find the point.

20 Tangent line problems Find an equation of the tangent line to f (x) = 3x 2 2 that is perpendicular to y = 2x + 5. Again, we need a point and a slope. Since we want a tangent line that is perpendicular to y = 2x + 5, which has a slope of 2. Since perpendicular lines have negative reciprocal slopes, we want a slope of 1 2. So we have our slope, in this problem we have to find the point. To find the point, we need to determine where on the graph of f (x) = 3x 2 2 is there a tangent line with slope 1 2. How do we do this?

21 Tangent line problems Find an equation of the tangent line to f (x) = 3x 2 2 that is perpendicular to y = 2x + 5. Again, we need a point and a slope. Since we want a tangent line that is perpendicular to y = 2x + 5, which has a slope of 2. Since perpendicular lines have negative reciprocal slopes, we want a slope of 1 2. So we have our slope, in this problem we have to find the point. To find the point, we need to determine where on the graph of f (x) = 3x 2 2 is there a tangent line with slope 1 2. How do we do this? Solve f (x) = 1 2!

22 Tangent line problems Find an equation of the tangent line to f (x) = 3x 2 2 that is perpendicular to y = 2x + 5. f (x) = 6x, so we need to solve 6x = which gives us x = 12. This is the x-value of the point on the graph of f (x) = 3x 2 2 that has a tangent line with slope 1 2.

23 Tangent line problems Find an equation of the tangent line to f (x) = 3x 2 2 that is perpendicular to y = 2x + 5. f (x) = 6x, so we need to solve 6x = which gives us x = 12. This is the x-value of the point on the graph of f (x) = 3x 2 2 that has a tangent line with slope 1 2. ( ) ( ) 2 To get the y-value of the point, f 1 12 = = Then the equation of the line is, y + 95 ( 48 = 1 x + 1 ). 2 12

24 Tangent line problems Determine the points (if any) on the graph of f (x) = x 3/2 2x where f has a horizontal tangent line.

25 Tangent line problems Determine the points (if any) on the graph of f (x) = x 3/2 2x where f has a horizontal tangent line. First, think about what type of slope a horizontal line has. If you can t remember draw a horizontal line, then choose two points on the line and use them to calculate the slope.

26 Tangent line problems Determine the points (if any) on the graph of f (x) = x 3/2 2x where f has a horizontal tangent line. First, think about what type of slope a horizontal line has. If you can t remember draw a horizontal line, then choose two points on the line and use them to calculate the slope. Horizontal lines have slope 0. This problem is asking if f (x) = 0 has any solutions. Calculate f (x) then click next.

27 Tangent line problems Determine the points (if any) on the graph of f (x) = x 3/2 2x where f has a horizontal tangent line. First, think about what type of slope a horizontal line has. If you can t remember draw a horizontal line, then choose two points on the line and use them to calculate the slope. Horizontal lines have slope 0. This problem is asking if f (x) = 0 has any solutions. Calculate f (x) then click next. f (x) = 3 2 x 1/2 2, now we solve f (x) = 0.

28 Tangent line problems Determine the points (if any) on the graph of f (x) = x 3/2 2x where f has a horizontal tangent line. 3 2 x 1/2 2 = 0 3 x = x = 3 x = 16 9 There ( is one ) point on f (x) that has a horizontal tangent line, 16 9, (To get y-value, plug x = 16 9 into f (x) = x 3/2 2x.)

29 3.3 Differential Equations Finally, let s go over differential equations. A differential equation is an equation that involves an unknown function and its derivative.

30 3.3 Differential Equations Finally, let s go over differential equations. A differential equation is an equation that involves an unknown function and its derivative. For example, y + xy = 3 is a differential equation; y is the unknown function and y is the derivative of the unknown function. We could also write this as f (x) + xf (x) = 3, but this does not look as nice and this is not traditionally how we write differential equations.

31 3.3 Differential Equations Finally, let s go over differential equations. A differential equation is an equation that involves an unknown function and its derivative. For example, y + xy = 3 is a differential equation; y is the unknown function and y is the derivative of the unknown function. We could also write this as f (x) + xf (x) = 3, but this does not look as nice and this is not traditionally how we write differential equations. Some more examples of differential equations are, y + 2y y = 0 y + y = 5 x 3 y + 2xy + y = 0

32 3.3 Differential Equations Given a differential equation, y + xy = 3, the goal is to find the unknown equation y. To do this we employ a variety of techniques from Calculus. There is a whole class devoted to this, Math3328 Differential Equations (which has Calculus III as a pre-requisite).

33 3.3 Differential Equations Given a differential equation, y + xy = 3, the goal is to find the unknown equation y. To do this we employ a variety of techniques from Calculus. There is a whole class devoted to this, Math3328 Differential Equations (which has Calculus III as a pre-requisite). We will not be finding the unknown function y. Now that you know how to differentiate a function, you can verify that a function y is a solution to a particular differential equation (and this is what we will do in the next example).

34 3.3 Differential Equations Verify that y = 3 cos(x) + sin(x) is a solution to y + y = 0. First we need to calculate y (do this before hitting next).

35 3.3 Differential Equations Verify that y = 3 cos(x) + sin(x) is a solution to y + y = 0. First we need to calculate y (do this before hitting next). y = 3 sin(x) + cos(x) y = 3 cos(x) sin(x)

36 3.3 Differential Equations Verify that y = 3 cos(x) + sin(x) is a solution to y + y = 0. First we need to calculate y (do this before hitting next). y = 3 sin(x) + cos(x) y = 3 cos(x) sin(x) Now we plug y and y into the equation. So, y + y = ( 3 cos(x) sin(x)) + (3 cos(x) + sin(x)) = 3 cos(x) + 3 cos(x) sin(x) + sin(x) = 0

37 3.3 Differential Equations Verify that y = 3 cos(x) + sin(x) is a solution to y + y = 0. First we need to calculate y (do this before hitting next). y = 3 sin(x) + cos(x) y = 3 cos(x) sin(x) Now we plug y and y into the equation. So, y + y = ( 3 cos(x) sin(x)) + (3 cos(x) + sin(x)) = 3 cos(x) + 3 cos(x) sin(x) + sin(x) = 0 Hence y = 3 cos(x) + sin(x) is a solution to y + y = 0.