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## CALCULUS I

by: Cassidy Grimes

24

0

4

# CALCULUS I MATH 141

Cassidy Grimes

GPA 3.51

Staff

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COURSE
PROF.
Staff
TYPE
Class Notes
PAGES
4
WORDS
KARMA
25 ?

## Popular in Mathematics (M)

This 4 page Class Notes was uploaded by Cassidy Grimes on Monday October 26, 2015. The Class Notes belongs to MATH 141 at University of South Carolina - Columbia taught by Staff in Fall. Since its upload, it has received 24 views. For similar materials see /class/229545/math-141-university-of-south-carolina-columbia in Mathematics (M) at University of South Carolina - Columbia.

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Date Created: 10/26/15
Problem 1 Use the limit de ntition of the derivative to nd when 12 Solution1 fzh7fz 1h2712 122zhh2712 f of 0 h2z h h 0 h fz limhno 239th 239th 239th limhno2z h 21 Problem 2 limxn z 7 7r cotz Solution 2 The limit is of the form 0 96 00 z 7 7r 1 7 7r 139 EHW 7 t l39 HW7Z39 xawi 2m 1 7r co 2m 1 WWI 2m tanltxgt Now the limit is of the form So we can apply L7Hopital7s Rule z 7 7r 1 272an lzm nwm 27217an cos2z 712 1 Problem 3 sin2z sin51 2771170 Solution 3 The limit is of the form so we can use L7Hopital7s Rule sin2z sin51 2cos2z 7 2 1 Scos51 7 5 1 liming liming 25 Problem 4 Find m L39 Solution 4 f 7 8662 96 12 7 tanz 21 7 z 96 8662 7 2tanz I 7 12 2 7 1 Problem 5 7r sin2zdz 0 Solution 5 Let u 21 then du 2dr So 07r sin2zdz 12 07r sinudu 127cosu 712 cos2zlzr 712 cos27r7712cos0 71212 0 Problem 6 Find 3 arcsin2z Solution 6 6 W 3lt 1 2 x 1 7 21 Problem 7 63x 17711700172 Solution 7 The limit is of the form so we can use L7Hopitalls rule 63x 363x 1771170072 hmsz z 21 This limit is also of the form so we can use LlHopitalls rule again l 363 7 l 963 7 7 27nme 21 7 27nme 2 7 2 7 00 Problem 8 Find 27 3 tanzy Solution 8 Take the derivative of both sides 0 se02zy 9 0 se02zyz g y 0 se02zy z 96 7 sec2zy y dy 8602Iy y dy dy 7y 7 2 2 i i i 7 sec 96 y sec 96 z 96 dx 8602zy I dz dz I Problem 9 lim 31 T 5 7H 61 7 8 Solution 9 The limit is of the form so we can use L7Hopitalls rule 31 5 3 lzm nm6x 7 8 2721754006 12 Problem 10 lim 21 8 EH74 12 z 712 Solution 10 2x 2z 4 2 2 l n l n l n 727 my 4z2z712 my 4173z4 my 4173 7473 Problem 11 12z3 7 75dz Solution 11 Let u 13 7 7 then du 312dz So lSdu IQdI Thus 2 37 5 7 5 7116713 76 zz 7dz713udu7367 18 C Problem 12 f is said to be continuous at z 5 provided that 3 conditions are satis ed What are the 3 conditions Solution 12 1 is de ned 2 lim nc z exists 3 lim nc z Problem 13 zg 7 4x2 ehdz IS 7 4x2 621d1 IS 7 412dz 21d1 Now let u 21 so du 2dr for the last integral Then Solution 13 4 3 4 3 4 3 3L 2 2x LL L u LL L uLL L 2x I 4zdze dz 4 4312edu 4 4312e 4 4312e C Problem 14 Find the local extrema of 3 fz7z27812 Solution 14 First we need to nd the critical points of The critical points are where 0 312 2 7 8 i 2 7 z 7 8 I 7 4z 2 So 0 when I 4 and z 72 The critical points are 4 and 72 Now we need to use either the rst derivative test or the second derivative test to determine if the critical points are local max or min 0 First Derivative Test Chose a number lower than 727 I chose 73 f73 7 gt 0 So is increasing from 00 to 72 Now chose a number between 72 and 47 I chose 0 fO 78 lt 0 So is decreasing from 72 to 4 Then chose a number greater than 47 I chose 5 f 5 7 gt 0 So is increasing from 4 to 00 Therefore has a local maximum at z 72 and a local minimum at z 0 Second Derivative Test f77 2x 7 2 f77 72 78 lt 0 Thus has a local maximum at z 72 f77 4 6 gt 0 Thus has a local minimum at z 4 Problem 15 Find 1W 111121 8Sl Solution 15 Using properties of logrithms7 we have that ln2z 7 83 3ln2z 7 8 So then 2 6 3 7 7 7 7 30732178 2178 3574 Problem 16 Find 161 2x new Solution 16 You can use either the product rule7 or in to simplify the expression and then implicit differentiation I Will use in y 2z Dew lny 1n2z 1 312 lny 1n2z 1 1n6312 lny 1n2z 1 3z2 Then after differentiating both sides we have that 1d yi6xgt dz dy Q 2z1 wwy i dz d e l e312 2 1212 6z dz 2 2 dy 2 2 7 6 2 1 3 i23 6 2 1 3 211 2I1 z e dz e z z e

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