Question

http://www.mastermathmentor.com/mmm/Free.aspx?bin=calc.AB%20Manual&file=AB39%20Calculus%20of%20e.pdf

@jim_thompson5910

This is gonna be what's next when you return

Answer 1

I did 1-12 but 9 I'm not 100% sure about

1. \[5e ^{5x}\]

2. \[-8 e ^{1-2x}\]

3. \[2x-3e ^{x^2-3x-1}\]

4. \[\frac{ e ^{\sqrt{x}} }{ 4\sqrt{x} }\]

5. \[2e^x(e^x+3)^2\]

6. \[e^x+xe^x\]

7. \[cose^x\]

8. \[cosxe ^{sinx}\]

9. \[\frac{ 1 }{ x } + 1\]

10. \[\frac{ 4xe ^{4x} - e ^{4x} }{ x^2 }\]

11. \[\frac{ e ^{4x}-4xe ^{4x} }{ (e ^{4x})^2 }\]

12. \[\frac{ e^x }{ 3(e^x)^{2/3} }\]

Answer 2

# 3 is incorrect, it should be

\[\Large (2x-3)e^{x^2 - 3x -1}\]

take note of the parenthesis around 2x-3

Answer 3

Thats what I have

Just forgot the ()

Answer 4

# 4 is incorrect, it should be

\[\Large \frac{e^{\sqrt{x}}}{\sqrt{x}}\]

the 2's will cancel

Answer 5

the initial 2 and the 2 that comes about when you derive sqrt(x)

Answer 6

Got it

Answer 7

#5, you won't have an exponent of 2 over the (e^x+3) since you pulled that exponent down (and you subtract 1 from 2 to get 2-1 = 1)

Answer 8

#6 is correct

Answer 9

#7 you forgot to do the chain rule

Answer 10

7 is e^xcose^x right?

Answer 11

yes \[\Large e^x*\cos(e^x)\]

Answer 12

Got it and 8 and 9?

Answer 13

you'll use the chain rule for 8

think of e^( sin(x) ) as just e^x for a temporary basis

Answer 14

then you'll multiply by the derivative of sin(x)

Answer 15

so...

y = e^( sin(x) )

dy/dx = cos(x)*e^( sin(x) )

Answer 16

Thats what I have :P

Answer 17

sweet

Answer 18

how about 9

Answer 19

And 9? I used the log rule

ln x and ln e^x

So 1/x and doesnt ln *e cancel leaving x and the derivative is 1?

Answer 20

you can't use any log rules here

Answer 21

you'd use the chain rule here as well

Answer 22

Isn't is 1+e^x/x+e^x then?

Answer 23

10 and 11 i did quotient and 12 i did chain,

after you check those we can skip over to 17 where integration starts :(

Answer 24

#9 is

\[\Large \frac{1+e^x}{x+e^x}\]

so you are correct

Answer 25

what did you get for 10?

Answer 26

I put it up above I put all of them up above lol

Answer 27

oh right, my bad lol, one sec

Answer 28

10 and 11 are correct

Answer 29

Not 12?

Answer 30

12 you can simplify to get \[\Large \frac{e^{x/3}}{3}\] however your answer is correct still

Answer 31

Oh okay cool

Answer 32

Now onto integration ew teach meee lol

And then after this just 1 extended response problem for my test corrections and we're good lol

Answer 33

for 17 through 20, you will use u-substitution to simplify the integrals

Answer 34

#17

let u = 5x

du/dx = 5 ---> du = 5*dx ---> dx = du/5

Answer 35

1/5 e^5x + c

Answer 36

correct

Answer 37

for #18, what can we make 'u' equal to?

Answer 38

It's -4e^1-x

Answer 39

-4e^(1-x), yes

Answer 40

And 19 is just e^x^2

Answer 41

yes, e^(x^2)

Answer 42

20 idk lol

Answer 43

let u = 1/x

du/dx = -1/(x^2) ---> du = -dx/(x^2) ---> -du = dx/(x^2)

Answer 44

-1 e^1/x?

Answer 45

close but no

Answer 46

you left out that 1/2

Answer 47

Oh right

Answer 48

Next is -e^sinx

Answer 49

it's positive though

Answer 50

double check your steps

Answer 51

hopefully you made u = sin(x)

Answer 52

Yeah but the integral of sin x is -cos x

Answer 53

u = sin(x)

du/dx = cos(x)

du = cos(x)dx

Answer 54

Oh duh sorry

Answer 55

\[\Large \int \cos(x)e^{\sin(x)}dx = \int e^{\sin(x)}*\cos(x)dx\]

\[\Large \int \cos(x)e^{\sin(x)}dx = \int e^{u}*du\]

Answer 56

its ok, we're going forwards and backwards between differentiating and integrating

Answer 57

I mix up the two a lot myself

Answer 58

Yeah ahahah

Okay so for 22 idk cuz i got 1/4 integral 1/u * du

and u^-1 when you integrate it goes away

Answer 59

Oh wait jk ln

Answer 60

Or no? Idk XD

Answer 61

it looks like you made u =4-e^x, correct?

Answer 62

Oh wait hangon

Answer 63

I can pull out the 1/4 and have

-1/4 integral of 1/u du still

Answer 64

I don't know where you're getting the 1/4 from

Answer 65

I pulled it out

Answer 66

This is the same as like 1/4 - e^x/e^x kinda thing

Answer 67

Oh nvm idk what I'm doing

Answer 68

ANYWAYSSSSS

u=4-e^x lol

Answer 69

-1 integral 1/u du now lol

Answer 70

u=4-e^x
du/dx = -e^x
-du = e^x*dx

\[\Large \int \frac{e^{x}}{4-e^{x}}dx = \int \frac{1}{4-e^{x}}*e^{x}dx\]
\[\Large \int \frac{e^{x}}{4-e^{x}}dx = \int \frac{1}{u}*(-du)\]
\[\Large \int \frac{e^{x}}{4-e^{x}}dx = -\int \frac{1}{u}du\]

Answer 71

yeah you got it

Answer 72

Okay but now what

You said i cant use the ln rule so

Answer 73

you can here because the integral of 1/u is ln|u|

Answer 74

I was referring back to when you had u^{-1/2} or something like that

Answer 75

Yay

-ln |4-e^x| +c

Answer 76

correct

Answer 77

btw, this packet is confusing because it stops at 24 on this page and then starts up again at 19 on the next page

Answer 78

Next is ln |e^x + e^-x| + c

Answer 79

you nailed it

Answer 80

24... has to be done in parts no?

Answer 81

yeah break up the fraction and put it into smaller pieces

Answer 82

its always a good idea to simplify anything you can before deriving or integrating

Answer 83

also keep in mind that 1/(e^x) = e^(-x)

Answer 84

Yeah idk how to do it lol

Answer 85

did you get this far?
\[\Large \int \frac{e^{4x}+2e^x+1}{e^x}dx = \int\left(\frac{e^{4x}}{e^x}+\frac{2e^{x}}{e^x}+\frac{1}{e^x}\right)dx\]
\[\Large \int \frac{e^{4x}+2e^x+1}{e^x}dx = \int\left(e^{3x}+2+e^{-x}\right)dx\]

Answer 86

Oh shoot no lol duh

Answer 87

1/3 e^3x +2x + e^-x?

Answer 88

+ C! lol

Answer 89

close, the integral of e^(-x) is -e^(-x) + C

Answer 90

so the overall answer is (1/3)*e^(3x) + 2x - e^(-x) + C

Answer 91

ok so what do you have for part a)?

Answer 92

Part a I got g(2) right which is 0.25

But I got g'(2) wong I think. I had -.05

Answer 93

g(2) should be negative since the area is under the x-axis