needing help on implicit differentiations?

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needing help on implicit differentiations?

Mathematics
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i can help
Really!
Can i write the equation down and you show me step by step how to do it?

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Yes
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)
(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
remember dy/dx e^(f(x)) = f'(x)e^(f(x))
okay so on the first line why are you taking out the Y in 2y?
oh are you taking y'
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
implicit*
for example d/dx 3 y^3 = 9 y ^2 dy/dx
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
sure
awesome here it is. Again thank you so much. I am in college and math is not my thing.
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)
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
3-1?
was that a typeo shouldnt it be 3-2/6-5=1?
or am i thinking wrong?
oops yep 1
oh good i was hoping i was starting to understand. could you help me correct the rest of my test?
(3-2)/(6-5)=1
sure
okay the next one i only got one point off but i am not really sure why here is the problem
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?
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
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.
she circled the +20 and put an arrow to the dx/dt
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
oh okay i am with you now.
600* 20 = 650 dz/dt dz/dt = 18.46
sorry the computer is slow today
no it is okay i am just so greatful for the help
:)
wow 20 is the answer.... what did the teacher want an extimate
idk maybe she only counted one off because i worked it mostly right just got the ending off
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
lol ok
okay so i had this idea so i dont have to type out all this and stuff i would just attach a file
that would be better
i am working on it give me a minute or so
here is the next page
1 Attachment
here is page 2
1 Attachment
here is the last page
1 Attachment
did they come in?
lol I need to download the new adobe sofeware
ha ha ha could you please
no can't see them
okay try this one
1 Attachment
1 Attachment
1 Attachment
is that better
NO cant read it don't know why
did you check the ones i just put up their in word
what microsoft word do you have?
because i can save it in a different format
try this
1 Attachment
it worked
okay awesome here come the rest
1 Attachment
1 Attachment
the order is 10,20,30
okay could we just go step by step and explain what i did wrong?
or is that too much?
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
awesome could we start at number 12 though
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
is that for 17?
yeah
thank you so much
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
okay tht makes since
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.
find the point of inflection.. f''(x) to zero and solve
going to bed I will help tomorrow if you need it?
okay how can i get a hold of you
i will post more ....you just log on. when do you need to hand it in?
by next monday but i have 5 more sections of homework i need to do also for the next sections
could i get help with those if needed? i can work on them tomorrow and then ask for help
ok

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