First we will go over the following derivative rule. Theorem

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1 Tuesday, Feb 1 Tese slides will cover te following 1 d [cos(x)] = sin(x) iger-order derivatives 3 tangent line problems 4 basic differential equations First we will go over te following derivative rule Teorem d [cos(x)] = sin(x) To see wy tis is true, we use te limit definition of te derivative Let f (x) =cos(x) and we ave f f (x + ) f (x) (x) cos(x + ) cos(x) Here we need to use te trigonometric identity cos(a + B) =cos(a) cos(b) sin(a)sin(b) wita = x and B = Tis means tat cos(x + ) =cos(x) cos() sin(x)sin() So we get, f cos(x + ) cos(x) (x) cos(x) cos() sin(x)sin() cos(x) (cos(x) cos() cos(x)) sin(x)sin() ( ( ) cos(x) cos() cos(x) ( cos(x) cos() cos(x) ( cos(x)(cos() 1) sin(x)sin() ) ( ) sin(x)sin() ) lim ) lim sin(x) ( ) sin()

2 Since tere are no s in sin(x) or cos(x), we can pull tem outside te limit like tis, [ ] [ ] cos() 1 sin() cos(x) lim sin(x) lim sin(x) Using our special trigonometric limits lim x 0 x =0,weget lim x 0 cos(x) 1 x [ cos(x) lim ] [ cos() 1 sin(x) lim and ence f (x) = sin(x) =1and ] sin() = cos(x)[0] sin(x)[1] = sin(x) Higer-order derivatives First consider f (x) =x 5 5x +sin(x) Now differentiate and we get f (x) =5x 4 10x + cos(x) wic we call te first derivative of f (x) If we differentiate f (x), tenweget f (x) =0x 3 10 sin(x) wic we call te second derivative of f (x) And if we differentiate tis again we get f (x) =60x cos(x) wic we call te tird derivative of f (x) Second, tird, fourt, etc derivatives are called iger-order derivatives Higer-order derivatives Te following is a cart of te notation we use for iger order derivatives function y y f (x) f (x) first derivative y dy f d (x) [f (x)] second derivative y d y f d (x) [f (x)] tird derivative y d 3 y 3 f (x) d 3 3 [f (x)] fourt derivative y 4 d 4 y 4 f 4 (x) d 4 4 [f (x)] fift derivative y 5 d 5 y 5 f 5 (x) d 5 5 [f (x)]

3 You will find lots of practice problems and omework problems tat simply ask you to differentiate Te following examples are to illustrate some of te types of tangent line problems tat you may come across First, let s take a look at a straigtforward tangent line problem Find an equation of te tangent line to f (x) = x +3at x =4 To write te equation of te tangent line we will need a point and aslope point: Atx =1,f (4) = = 5, and so te point is (4, 5) slope: We differentiate to get f (x) = 1 x 1/,andten m = f (4) = 1 ( 1 )( 1 ) ( 1 )( 1 (4) 1/ = 4 = = ) 1 4 Terefore te equation of te tangent line is y 5= 1 (x 4) 4 Find an equation of te tangent line to f (x) =3x tat is perpendicular to y =x +5 Again, we need a point and a slope Since we want a tangent line tat is perpendicular to y =x + 5, wic as a slope of Since perpendicular lines ave negative reciprocal slopes, we want a slope of 1 Soweaveourslope,intisproblemweavetofindte point To find te point, we need to determine were on te grap of f (x) =3x is tere a tangent line wit slope 1 Howdowe do tis? Solve f (x) = 1!

4 Find an equation of te tangent line to f (x) =3x tat is perpendicular to y =x +5 f (x) =6x, soweneedtosolve6x = 1 1 wic gives us x = 1 Tis is te x-value of te point on te grap of f (x) =3x tat as a tangent line wit slope 1 ( ) ( ) To get te y-value of te point, f 1 1 =3 1 1 = Ten te equation of te line is, y + 95 ( 48 = 1 x + 1 ) 1 Determine te points (if any) on te grap of f (x) =x 3/ x were f as a orizontal tangent line First, tink about wat type of slope a orizontal line as If you can t remember draw a orizontal line, ten coose two points on te line and use tem to calculate te slope Horizontal lines ave slope 0 Tis problem is asking if f (x) =0 as any solutions Calculate f (x) tenclicknext f (x) = 3 x 1/, now we solve f (x) =0 Determine te points (if any) on te grap of f (x) =x 3/ x were f as a orizontal tangent line 3 x 1/ =0 3 x = 4 x = 3 x = 16 9 Tere ( is one ) point on f (x) tat as a orizontal tangent line, 16 9, 3 7 (To get y-value, plug x = 16 9 into f (x) =x 3/ x)

5 33 Differential Equations Finally, let s go over differential equations A differential equation is an equation tat involves an unknown function and its derivative For example, y + xy =3is a differential equation; y is te unknown function and y is te derivative of te unknown function We could also write tis as f (x)+xf (x) =3, but tis does not look as nice and tis is not traditionally ow we write differential equations Some more examples of differential equations are, y +y y =0 y + y =5 x 3 y +xy + y =0 33 Differential Equations Given a differential equation, y + xy = 3, te goal is to find te unknown equation y To do tis we employ a variety of tecniques from Calculus Tere is a wole class devoted to tis, Mat338 Differential Equations (wic as Calculus III as a pre-requisite) We will not be finding te unknown function y Nowtatyou know ow to differentiate a function, you can verify tat a function y is a solution to a particular differential equation (and tis is wat we will do in te next example) 33 Differential Equations Verify tat y =3cos(x)+sin(x) is a solution to y + y =0 First we need to calculate y (do tis before itting next) y = 3sin(x)+cos(x) y = 3 cos(x) sin(x) Now we plug y and y into te equation So, y + y =( 3cos(x) sin(x)) + (3 cos(x)+sin(x)) = 3 cos(x) + 3 cos(x) sin(x)+sin(x) =0 Hence y =3cos(x)+sin(x) is a solution to y + y =0