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mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0@Paynesdad It's posted ^_^

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0am i able to use the chain rule for this problem?

Paynesdad
 one year ago
Best ResponseYou've already chosen the best response.1Ok 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.

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

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

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0im getting there.

Paynesdad
 one year ago
Best ResponseYou've already chosen the best response.1no 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.

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

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

mebs
 one year ago
Best ResponseYou've already chosen the best response.1@Paynesdad I think this is implicit because they wanted it in terms of Q.

Paynesdad
 one year ago
Best ResponseYou've already chosen the best response.1Seems like you answered it mebs...not helped with it.

mebs
 one year ago
Best ResponseYou've already chosen the best response.1@Paynesdad I think your right....my apologies I get exited when I do math.

Paynesdad
 one year ago
Best ResponseYou've already chosen the best response.1LOL I know the feeling! I am glad I am not the only one.

mebs
 one year ago
Best ResponseYou've already chosen the best response.1Well @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.

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0sorry just doing the math

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0yes i see what you mean.

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0im not up to the answer yet but close, i just need to check

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0i got 48.96.... :? but it's not right..

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0would using the chain rule be wrong?

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0wait, which one's?

mebs
 one year ago
Best ResponseYou've already chosen the best response.1You can't do what you just did in your 2nd and third step

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0oh no, :? why not?

Paynesdad
 one year ago
Best ResponseYou've already chosen the best response.1Whoa whoa whoa.. Hang on here.

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

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0oh okay, i understand that.

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0i used it twice before.

Paynesdad
 one year ago
Best ResponseYou've already chosen the best response.1Remember 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.

mebs
 one year ago
Best ResponseYou've already chosen the best response.1You 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.

Paynesdad
 one year ago
Best ResponseYou've already chosen the best response.1So using mebs equation you should get...a fractional answer.

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0gotta start over, i'll keep you posted.

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

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0damn it. wrong again. i got 2/33 this time

Paynesdad
 one year ago
Best ResponseYou've already chosen the best response.1Show us what you did again if you can.

Paynesdad
 one year ago
Best ResponseYou've already chosen the best response.1Got it...Order of operations!

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0where did i miss?

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

mebs
 one year ago
Best ResponseYou've already chosen the best response.1You can't add them they both don't have dy/dx dude.

Paynesdad
 one year ago
Best ResponseYou've already chosen the best response.1you 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

mebs
 one year ago
Best ResponseYou've already chosen the best response.1you get \[2 = 12 + 45\frac{dy}{dx}\]

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0ohhhhh 45 is part of dy/dx

mebs
 one year ago
Best ResponseYou've already chosen the best response.1yea bro, step It up man dude.

Paynesdad
 one year ago
Best ResponseYou've already chosen the best response.1lol yep... it is a coefficient...come on bro!

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0i wasn't looking closely.

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0p.s not a bro lol but thanks! ^_^

mebs
 one year ago
Best ResponseYou've already chosen the best response.1Its fine I was jking haha.hhaa

mebs
 one year ago
Best ResponseYou've already chosen the best response.1I can tell that by your handwriting haha.haha

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0kay, well thanks guys!!! appreciate it tons tons.

mebs
 one year ago
Best ResponseYou've already chosen the best response.1Good question @mathcalculus keep up the hard work.

Paynesdad
 one year ago
Best ResponseYou've already chosen the best response.1@mebs you crack me up

Paynesdad
 one year ago
Best ResponseYou've already chosen the best response.1Keep it up @mathcalculus
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