Please do all the even numbers like 34, 36, 38, 40,42, 44, 46, 48, 50, 52, 54, 56, 58, 60, 62.. Please do it on paper bc it's really hard to understand when is written on here. Also, please show step by step how you got that answer even if there's a simplifying part. I'm gonna be attaching some examples and it would be really helpful if you follow the same format so is more easy for me to understand.

Instructions: Differentiate

You can use the chain rule or product rule

34. y = 2x + 3 V3x - 5 3x - 4 35. y =V 5x + 3 36. f(x) = tan(xVx - 1) 37. r(x) = x(0.01x2 + 2.391x - 8.51)S 38. r(x) = (3.21x - 5.87)3(2.36x - 5.45)5 39. y = Vcot 5x - cos 5x 40. y = cot x - sin(cos2x) 41. y = sin(sec*(x2)) 42. f(x) = VV2x + 3+1 43. 8(x) = VV x2 + 2 +1 X 44. y= (x + sin x)2 sin x 45. y = x2 + 5 46. y = tan (Vt + 2) 47. f(x) = cot3(x sin(2x + 4))48. y = V2 + cos2t 49. y = Vsectx + x 50. y = Vx + csc x Find ay du du ' dy . and ax dx 51. y = Vu and u = x2 - 1 15 52. y = =3 and I = 2x + 1 53. y = u30 and u = 4x3 - 2x2 u+ 1 54 . y = 1 - 1 and u = 1 + Vx 55. y = u(u + 1) and u = x3 - 2x 56. y = (u + 1)(u - 1) and u = x3 + 1 57. Find an equation for the tangent line to the graph of y = Vx2 + 3x at the point (1, 2). 58. Find an equation for the tangent line to the graph of y = (x3 - 4x)10 at the point (2, 0). 59. Find an equation for the tangent line to the graph of y = x V2x + 3 at the point (3, 9). 50. Find an equation for the tangent line to the graph 2x + 3 \\3 of y = x - at the point (2, 343). 61. Find an equation for the tangent line to the graph of f(x) = sin x at the point (-7/6, 1/4). 62. Find an equation for the tangent line to the graph of f(x) = x sin 2x at the point (77, 0).\fDate No 3 ) y = (1 - 2 x ) 55 when the Chain Rule # just have F = 1 - 2 x ( ) = - 2 16 an X and not K = 55 y'= 55 [ 1-2x ] 5 - ( -2 ) exponent then is just Y = - 10 [1- 2x 754 the I 2 x = 2 5 ) y = sec 2 x - 110 [ 1- ( x ] Rewrite y = ( sec x ) 2 Rule 5 f = secx J F - secxtanx ( 1 ) 2 - 1 15 = 2 y = a [ secx ] ' ( secx tanx ) 5 = 1/ 2 7 1= 1 1 - 3 x = ( 1- 3)/2 F = 1 - 3 x F ) = - 3 15 = 1/2 dy = 1 [ 1 - 3 x]12 -' 1 - 37 dx' = = 3 ( 1- 3 x) 12 if you don't " ) f ( x ) _- 2 Two ways see exponent 3 x2 + 1 then is - 1 chain Rule Rewrite : 2 (3 x2 + 7 ) - * f = 3 x 2 + 1 ( ) = 3 ( 2 ) x " = 6 x 19 = - 1 1-1 2 - 2 2 . - 1 . 6 = F ) ( x ) = 2 ( - 1) ( 3 x 2 +1 ) ( 6 x ) = 12 x (3x2 + 4 ) - 2 mo\fDate No * 10 ) g ( x ) = 2/2 x - 1 + ( 41 * x 2 ) = ( 2 x = 1713 + ( 41 - X ) ? F = 2 x = 1 f 1 - 2 1 9 = 4 - x 15 = 3 n = 2 27 simplify 9' = $ ( 2 x - 1) 213 ( 2 ) +2 ( 4 + x ) 1-1) Product R product * ally = x'sin x + 5 x Cos * + ( / sec x x is a function dy G Ex 2 3 x 'sinx + x 3 (osX + 5 (osX + 5x(- sinx )+ 4 seextank BIT31 x 23 ) y = V x 2 + x 3 (* 2 x?+ 3 x + 5 ) Product Rule = ( x 2 7 x 3 ) "/2 ( 2 x 2 + 3 x + 5 1 f = ( x ? + * 3 ) 1 / 2 F 1 = 2 ( x 2 4 x 3 ) 12 ( 2 * + 3 * 2) (2 = ( 2 x2 + 3 x +5 ) ( ) = 2 ( 2 ) x ' +3 G ! = 4 X + 3 G yo = 2 ( * ? + * 3 ) " ( 2 x +3x ? ) ( 2 x " + 3 * 45 ) + ( x 2+ * 3 ) 12 (14 * + 3)No Date Trigonometric function 29 ) y = sin (cosx ) U = COSX U' = - Sinx (os U U) dy = cos ( cosx ) ( sink ) -U 34) 4 3/ 2 x + 3 3 x - 5 = ( 2 x + 3 ) /3 Quotient ? = d [ c ].0' v - Uv. 3 x - 5 F . = 2 x + 3 ' " f ) = 2 ( 3 x - 5 ) - (2 x + 3) (3 ) 2 *+ 3 3 x - 5 V (3 x - 5)2 m : 2 1h = 3 3 x - 5 n = 3 y'= 3 22 x + 3 -2/3 L3x - 5 2 ( 3x - 5 - (24+ 3) (3) ( 3 x - 5 ) 2 Second way to do it y = ( ax+ 3) 1/3 ( 3 * - 5 ) 13 G F = (2 * +3 ) "3 ()=3 ( 2 *+3)-2/3 ( 2 ) ( = ( 3 x -5 ) " 3 41=1 (3x-5)-213 (3) 1 ' 3 3 ( 2 x 4 3 ) 3 ( 2 ) ( 3 x - 5 ) 13 - ( 2 * + 3 ) 13 3 ( 3 x- 5)1/ 3 3 [ ( 34 - 5) 13 72