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mathcalculusBest ResponseYou've already chosen the best response.0
@Paynesdad It's posted ^_^
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
am i able to use the chain rule for this problem?
 8 months ago

PaynesdadBest ResponseYou've already chosen the best response.1
Ok so take the derivative with respect of the equation using implicit differentiation. Since both x and y are variables that depend on t you will have to use the chain rule.
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
okay so i began doing this: DQ/DT= (2X)(5Y)
 8 months ago

mebsBest ResponseYou've already chosen the best response.1
\[\frac{dQ}{dt} = 2 = 2x\frac{dx}{dt}5y^{4}\frac{dy}{dt}\]
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
im getting there.
 8 months ago

PaynesdadBest ResponseYou've already chosen the best response.1
no you can't do the derivative of x and y and the same time. \[\frac{ dQ }{ dt }= \frac{ d(x^2y^5) }{ dt }\]First take the derivative of the right side with respect to t of the x factor and treat y factor as a constant + the derivative of the right side with respect to t of the y factor treating the x factor as a constant.
 8 months ago

mebsBest ResponseYou've already chosen the best response.1
we know that \[\frac{dx}{dt} = 2\] so when now find what \[\frac{dy}{dt} , at, x = 3, y = 1\]
 8 months ago

mebsBest ResponseYou've already chosen the best response.1
So solve this \[2 = 2(3)(2)5(1)^{4}\frac{dy}{dt}\]
 8 months ago

mebsBest ResponseYou've already chosen the best response.1
@Paynesdad I think this is implicit because they wanted it in terms of Q.
 8 months ago

PaynesdadBest ResponseYou've already chosen the best response.1
Seems like you answered it mebs...not helped with it.
 8 months ago

mebsBest ResponseYou've already chosen the best response.1
@Paynesdad I think your right....my apologies I get exited when I do math.
 8 months ago

PaynesdadBest ResponseYou've already chosen the best response.1
LOL I know the feeling! I am glad I am not the only one.
 8 months ago

mebsBest ResponseYou've already chosen the best response.1
Well @mathcalculus does this make sense to you.. I just subbed values that were given to me already from the question into the derivative of Q. and x and y were already defined.
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
sorry just doing the math
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
yes i see what you mean.
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
im not up to the answer yet but close, i just need to check
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
i got 48.96.... :? but it's not right..
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
would using the chain rule be wrong?
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
wait, which one's?
 8 months ago

mebsBest ResponseYou've already chosen the best response.1
You can't do what you just did in your 2nd and third step
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
oh no, :? why not?
 8 months ago

PaynesdadBest ResponseYou've already chosen the best response.1
Whoa whoa whoa.. Hang on here.
 8 months ago

mebsBest ResponseYou've already chosen the best response.1
To use the product rule you do one at a time.\[\frac{dQ}{dt} = 2x\frac{dx}{dt}y^{5} + x^{2}5y^{4}\frac{dy}{dt}\]
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
oh okay, i understand that.
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
i used it twice before.
 8 months ago

PaynesdadBest ResponseYou've already chosen the best response.1
Remember d(xy)/dt = y (dx/dt) + x (dy/dt). Also mathcalculus when you took the derivative you left off the dx/dt and dy/dt out of your equations.
 8 months ago

mebsBest ResponseYou've already chosen the best response.1
You already know that \[\frac{dQ}{dt}= 2 \] You know that \[\frac{dx}{dt} = 2\] and you know that \[x = 3 , y = 1\] just substitute it into the derivative of dQ/dt and solve.
 8 months ago

PaynesdadBest ResponseYou've already chosen the best response.1
So using mebs equation you should get...a fractional answer.
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
gotta start over, i'll keep you posted.
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
okay, jus wondering. so am i able to substitute all of them right away after i found the derivative?
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
damn it. wrong again. i got 2/33 this time
 8 months ago

PaynesdadBest ResponseYou've already chosen the best response.1
Show us what you did again if you can.
 8 months ago

PaynesdadBest ResponseYou've already chosen the best response.1
Got it...Order of operations!
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
where did i miss?
 8 months ago

mebsBest ResponseYou've already chosen the best response.1
Let me show you what must be done. \[2 = 2\times 3 \times (2) \times (1)^{5} + (3)^{2} \times 5 \times 1^{4} \frac{dy}{dx}\]
 8 months ago

mebsBest ResponseYou've already chosen the best response.1
You can't add them they both don't have dy/dx dude.
 8 months ago

PaynesdadBest ResponseYou've already chosen the best response.1
you have 12 + 45*dy/dt=2 ...You can't add 12 to 45 until you multiply 45 times dy/dt...It's like 4=3+7x....you don't simplify that as 4 = 4x ... right
 8 months ago

mebsBest ResponseYou've already chosen the best response.1
you get \[2 = 12 + 45\frac{dy}{dx}\]
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
ohhhhh 45 is part of dy/dx
 8 months ago

mebsBest ResponseYou've already chosen the best response.1
yea bro, step It up man dude.
 8 months ago

PaynesdadBest ResponseYou've already chosen the best response.1
lol yep... it is a coefficient...come on bro!
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
i wasn't looking closely.
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
p.s not a bro lol but thanks! ^_^
 8 months ago

mebsBest ResponseYou've already chosen the best response.1
Its fine I was jking haha.hhaa
 8 months ago

mebsBest ResponseYou've already chosen the best response.1
I can tell that by your handwriting haha.haha
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
kay, well thanks guys!!! appreciate it tons tons.
 8 months ago

mebsBest ResponseYou've already chosen the best response.1
Good question @mathcalculus keep up the hard work.
 8 months ago

PaynesdadBest ResponseYou've already chosen the best response.1
@mebs you crack me up
 8 months ago

PaynesdadBest ResponseYou've already chosen the best response.1
Keep it up @mathcalculus
 8 months ago
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