Question

needing help on implicit differentiations?

Answer 1

i can help

Answer 2

Really!

Answer 3

Can i write the equation down and you show me step by step how to do it?

Answer 4

Yes

Answer 5

okay this is the instructions...Use implicit differentiation to find the value of dy/dx at the indicated point.

e^y+2y=3x-5 ; at the point (2,0)

Answer 6

(dy/dx)e^y + 2dy/dx = 3
dy/dx(e^y + 2) = 3
(dy/dx)= 3/(e^y + 2)
plug in the value of x an y in the equation but there isn't a y varible
( dy/dx)= 3/(e^0 + 2)
( dy/dx)= 3/3 = 1

Answer 7

remember dy/dx e^(f(x)) = f'(x)e^(f(x))

Answer 8

okay so on the first line why are you taking out the Y in 2y?

Answer 9

oh are you taking y'

Answer 10

implicate differentiate is all about in term of a variable and the variable in this problem is x. So differentiate the polynomials and if DOESNT have x in the polynomial then add dy/dx to the polynomial

Answer 11

implicit*

Answer 12

for example d/dx 3 y^3 = 9 y ^2 dy/dx

Answer 13

oh okay. can you help me on another one? i am trying to correct a test and want to make sure i understand this before i attempt the next homework

Answer 14

sure

Answer 15

awesome here it is. Again thank you so much. I am in college and math is not my thing.

Answer 16

awesome here it is. Again thank you so much. I am in college and math is not my thing.

xy-2x=y ^{2} - 4 ; at the point (5,3)

Answer 17

xy-2x=y ^{2} - 4 ; at the point (5,3)
xdy/dx + y - 2 = 2ydy/dx /// remember the multiplication rule
dy/dx(2y - x)= y-2
dy/dx= (y-2)/((2y - x))
plug in values
(3-1)/(6-5)= 2

Answer 18

3-1?

Answer 19

was that a typeo shouldnt it be 3-2/6-5=1?

Answer 20

or am i thinking wrong?

Answer 21

oops yep 1

Answer 22

oh good i was hoping i was starting to understand.

could you help me correct the rest of my test?

Answer 23

(3-2)/(6-5)=1

Answer 24

sure

Answer 25

okay the next one i only got one point off but i am not really sure why here is the problem

Answer 26

a women standing on a cliff is watching a motorboat through a telescope as the boat travels away from the base of the cliff. The telescope is 250 feet above the water level and the boat is traveling at a speed of 20 feet per second. How fast is the distance between the woman and the boat increasing at the moment when the boat is 600 feet from away?

Answer 27

here is what i did

x^2 + 250^2 = z^2
2x + 0 = 2z
2x dx/dt = 2z dx/dt
600dx/dt=650(+20)
dx/dt = 21.7 ft/s

Answer 28

The boat and the women are in a triangle formation. So \[x^2 +y^2 = z^2\] implicit differentiate in terms of time. so \[(x) dx/dt + (y) dy/dt = (z) dz/dt\]
where x= the distance the boat is and dx/dt the rate of the boat moving away. because the women is sationary the women's dy/dt is 0.

Answer 29

she circled the +20 and put an arrow to the dx/dt

Answer 30

So (x)dx/dt=(z)dz/dt

600* 20 = z dz/dt
we are trying to find dz/dt but we don't have z
x^2+y^2=z^2

600^2 + 250^2 = z^2
z=650

Answer 31

oh okay i am with you now.

Answer 32

600* 20 = 650 dz/dt
dz/dt = 18.46

Answer 33

sorry the computer is slow today

Answer 34

no it is okay i am just so greatful for the help

Answer 35

:)

Answer 36

wow 20 is the answer.... what did the teacher want an extimate

Answer 37

idk maybe she only counted one off because i worked it mostly right just got the ending off

Answer 38

okay ready for the next one? just to warn you i have like 11 left i need help on, but they get easier i think..lol

Answer 39

lol ok

Answer 40

okay so i had this idea so i dont have to type out all this and stuff i would just attach a file

Answer 41

that would be better

Answer 42

i am working on it give me a minute or so

Answer 43

here is the next page

Answer 44

here is page 2

Answer 45

here is the last page

Answer 46

did they come in?

Answer 47

lol I need to download the new adobe sofeware

Answer 48

ha ha ha could you please

Answer 49

no can't see them

Answer 50

okay try this one

Answer 51

is that better

Answer 52

NO cant read it don't know why

Answer 53

did you check the ones i just put up their in word

Answer 54

what microsoft word do you have?

Answer 55

because i can save it in a different format

Answer 56

try this

Answer 57

it worked

Answer 58

okay awesome here come the rest

Answer 59

the order is 10,20,30

Answer 60

okay could we just go step by step and explain what i did wrong?

Answer 61

or is that too much?

Answer 62

number 17:
2x^3+ 3x^2 = 2 = f(x)
12x^2 + 6x = f'(x) = 0 at x =-1/2 and x= 0
24x+ 6 = f''(x) = 0 at -1/4

Answer 63

awesome could we start at number 12 though

Answer 64

increasing [-∞. -1/2][0, ∞]
decreasing [-1/4,-1/2]

local/absolute maximum when f'(x) = 0 and f''(x) < 0 at x=-1/2
local/absolute minimum when f'(x) = 0 and f''(x) > 0 at x=0

Answer 65

is that for 17?

Answer 66

yeah

Answer 67

thank you so much

Answer 68

find the intervals of decreasing and increasing take the derivative and set it to zero. the number plug it in the ORIGINAL f(x) and if the number is negative the between the local max or min it is decreasing and whent he number is positive between the local max or min it is increasing

Answer 69

okay tht makes since

Answer 70

to find the local max and min.... find f''(x) and plug in the f'(x) number and if it is negative (positive) it is a local maximum(minimum). of the local max and min plug them in the original f(x) AND which ever the local max or min is the most positive or negative is the absolute max or min respectively.

Answer 71

find the point of inflection.. f''(x) to zero and solve

Answer 72

going to bed I will help tomorrow if you need it?

Answer 73

okay how can i get a hold of you

Answer 74

i will post more ....you just log on. when do you need to hand it in?

Answer 75

by next monday but i have 5 more sections of homework i need to do also for the next sections

Answer 76

could i get help with those if needed? i can work on them tomorrow and then ask for help

Answer 77

ok