needing help on implicit differentiations?

- anonymous

needing help on implicit differentiations?

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- anonymous

i can help

- anonymous

Really!

- anonymous

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

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- anonymous

Yes

- anonymous

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)

- anonymous

(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

- anonymous

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

- anonymous

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

- anonymous

oh are you taking y'

- anonymous

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

- anonymous

implicit*

- anonymous

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

- anonymous

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

- anonymous

sure

- anonymous

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

- anonymous

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)

- anonymous

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

- anonymous

3-1?

- anonymous

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

- anonymous

or am i thinking wrong?

- anonymous

oops yep 1

- anonymous

oh good i was hoping i was starting to understand.
could you help me correct the rest of my test?

- anonymous

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

- anonymous

sure

- anonymous

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

- anonymous

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?

- anonymous

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

- anonymous

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.

- anonymous

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

- anonymous

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

- anonymous

oh okay i am with you now.

- anonymous

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

- anonymous

sorry the computer is slow today

- anonymous

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

- anonymous

:)

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

lol ok

- anonymous

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

- anonymous

that would be better

- anonymous

i am working on it give me a minute or so

- anonymous

here is the next page

- anonymous

##### 1 Attachment

- anonymous

here is page 2

- anonymous

##### 1 Attachment

- anonymous

here is the last page

- anonymous

##### 1 Attachment

- anonymous

did they come in?

- anonymous

lol I need to download the new adobe sofeware

- anonymous

ha ha ha could you please

- anonymous

no can't see them

- anonymous

okay try this one

- anonymous

##### 1 Attachment

- anonymous

##### 1 Attachment

- anonymous

##### 1 Attachment

- anonymous

is that better

- anonymous

NO cant read it don't know why

- anonymous

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

- anonymous

what microsoft word do you have?

- anonymous

because i can save it in a different format

- anonymous

try this

##### 1 Attachment

- anonymous

it worked

- anonymous

okay awesome here come the rest

- anonymous

##### 1 Attachment

- anonymous

##### 1 Attachment

- anonymous

the order is 10,20,30

- anonymous

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

- anonymous

or is that too much?

- anonymous

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

- anonymous

awesome could we start at number 12 though

- anonymous

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

- anonymous

is that for 17?

- anonymous

yeah

- anonymous

thank you so much

- anonymous

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

- anonymous

okay tht makes since

- anonymous

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.

- anonymous

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

- anonymous

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

- anonymous

okay how can i get a hold of you

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

ok

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