MATH section 2.3 Basic Differentiation Formulas Page 1 of 5

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1 MATH 0100 section. Basic Dierentiation Formulas Page 1 o The tetbook is using Leibniz notation or this section. I ll continue to use this notation when we are using ormulas an rules. Notation: This means, take a erivative o the epression insie the parenthesis with respect to the _ variable in the enominator I let as an unerscore because the variable can change. y θ This means, take a erivative o the epression insie the parenthesis with respect to. This means, take a erivative o the epression insie the parenthesis with respect to t. This means, take a erivative o the epression insie the parenthesis with respect to y. This means, take a erivative o the epression insie the parenthesis with respect to θ. For simplicity: I a unction is eine as, then we can write I a unction is eine as y = I a unction is eine as p =, then we can write y, then we can write p or erivative o with respect to. or erivative o y with respect to. or erivative o p with respect to. The Derivative rules given below which are rom tet are written with an assumption that we are taking erivative with respect to. Derivative o a Constant Function Given c is any constant, then c = 0. y Or i y = c, then 0 = The Power Rule General Version n n 1 I n is any real number, then n = n y n 1 Or i y =, then = n Also, i n =1, then = = 1 = 1 = 1 The Constant Multiple Rule I c is a constant an is a ierentiable unction, then [ c ] = c

2 MATH 0100 section. Basic Dierentiation Formulas Page o The Sum Rule I an g are both ierentiable, then [ + g ] = + g The Dierence Rule I an g are both ierentiable, then [ g ] = g The Since an Cosine Functions sin = cos cos = sin Note: Use the Leibniz notation an pay attention to the variable in the enominator; this notation inorms you to take a erivative in respect to which variable. For eample: I y =, an the question is in y, then the correct answer is y = = 0 not because it was aske to take the erivative with respect to t not. Also, to make the notes easier to unerstan in uture I ll use square bracket when erivatives are applie. For y eample, y =, then = = 1. Higher Derivatives The rules given above are taking 1 step o a erivative. I we apply another roun o erivative rules to a erivative o a unction, then we call it n erivative. I we continue then we get r erivative an so on. Table below shows notation o both Leibniz an prime notation or higher erivatives. Given y = Derivative Orer Leibniz Notation Prime Notation 1 st y n y y = y y = r y y = th th y y = n n y y n 1 = n n n th 1

3 MATH 0100 section. Basic Dierentiation Formulas Page o Aitional eamples: 6 t = 1.t.t = 1. t.[ t] + 0 = 7t t 10 h = + h h [ ] = + = 6 = [1] 0= y = y 1 = = = SR = π R S = π [ R] = 8π R R 0 gu = u+ u 1 g 1 1 gu = u+ u= u+ u = [1] + u u = + u y = + y = y 1 = + = + = + = 1 + = = 6 8 u = t + t 1 1 u u = t + t = t + t = t t + = + t t y = 1, 0 y y = [ ] = y 0 = 1 Tangent line: mt = = 1 1 = = 1 y = + Normal line is the line that coming out straight out o the curve at speciic point or perpenicular to the tangent line at the same point: mn = = = y 0 = 1 y = m T

4 MATH 0100 section. Basic Dierentiation Formulas Page o ht = t+ sin t h h = t + cos = t sint sin ht t t + = = t + sint = + cost = sint t [ t] [ ] t 8 y = points where tangent line is horizontal y = + [ ] 1[1] + 0 = To in the values o that the tangent line is horizontal we nee to take the 1 st erivative we solve above, set it equal to 0 an solve or. y 0= = = = 0 1= 0 0= 6 + = = 1 To get the y coorinates o the points, or each value oun above, plus it into the original epression: At = : y = = + + = = an the point is,1 At = 1: y = = + + = + + = an the point is 1, y = equation o the tangent line is parallel to y = 1+ y = y 1 = = = = We must in the point on the graph where this conition occurs. The slope o the line is m =, so let the 1 st erivative we oun above equal to the value o the slope an then solve or. y = = 1= = = At = : y = = = = 8 an the point is,8 So the tangent line is y 8 = y 8 = 1 y= 8 v0 = m/s s= t+ t m a vt = s = [1] + [ t] = + 6t s v = = + 6 = + 1 = 17 m/s b vt = m/s? + 6t = 6t = 0 t = s

5 MATH 0100 section. Basic Dierentiation Formulas Page o 0 v0 = 80 t/s s= 80t 16t t this is with setting upwar as positive velocity an own as negative velocity. a At the maimum height, the velocity is zero, vt = 0t/s s 0= 80 t 80 vt = = 80[1] 16[ t] = 80 t = s t = 80 Now plug the result into the original equation to get the height: s = t t = = = = b Since the height o 96 t is less than our maimum height o 100 t, this will occur on its way up an own. So set the height equal to 96 t an solve or t. 96 = 80t 16t 16 t t = 0 t t+ = t = t = t t+ = t = t = s On its way up t = s with velocity v = = 80 = 80 6 = 16 t/s s On its way own t = s with velocity v = = 80 = = 16 t/s 1 V = t 0 t V 1 1 V = t = [1] + [ t] t 1 000t 000 t = t+ t = 000 = = V a at t = min, = 0 = = gallons/min V b at t = 10 min, = 0 = = = gallons/min 10 V c at t = 0 min, = 0 = = = gallons/min 0 V 0 at t = min, = 0 = 0 0 = 0 gallons/min 0 We see that as the tank empties, the rate is ecreasing an eventually, at t = min, the rate is 0 gallons/min. But just to be sure the computation o when this starts is calculate below to justiy our V 0 result that we have heist rate at the beginning, at t = 0min, = 0 = 0 gallons/min. 0

f sends 1 to 2, so f 1 sends 2 back to 1. f sends 2 to 4, so f 1 sends 4 back to 2. f 1 = { (2,1), (4,2), (6,3), (1,4), (3,5), (5,6) }.

f sends 1 to 2, so f 1 sends 2 back to 1. f sends 2 to 4, so f 1 sends 4 back to 2. f 1 = { (2,1), (4,2), (6,3), (1,4), (3,5), (5,6) }. .3 Inverse Functions an their Derivatives In this unit you ll review inverse unctions, how to in them, an how the graphs o unctions an their inverses are relate geometrically. Not all unctions can be unone,