anonymous
  • anonymous
needing help on implicit differentiations?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
i can help
anonymous
  • anonymous
Really!
anonymous
  • anonymous
Can i write the equation down and you show me step by step how to do it?

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

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