Question

Assume x and y are functions of t. Evaluate dy/dt (the derivative of y with respect to t) for 4xy-3x+5y = -40 given the conditions dx/dt=-9, x=3, y=-1. Why does the dy/dt of 4xy equal 4x(dy/dt) + 4y(dx/dt)? I know you have to use the product and chain rules and, I can get and understand the first term, but I don't understand how they get the 2nd term 4y(dx/dt).

Answer 1

just take an implicit with respect to t

Answer 2

4xy -3x+5y = -40

D[4xy] -D[3x] +D[5y] = D[-40]

4D[xy] -3D[x] +5D[y] = D[-40]

4D[xy] -3x' +5y' = 0

hmm, how do we do the product of xy ?

Answer 3

D[uv] = D[u] v + u D[v]
= u' v + u v'

Answer 4

4D[xy] -3x' +5y' = 0

4(D[x] y + x D[y]) -3x' +5y' = 0

4( x' y + x y') -3x' +5y' = 0

Answer 5

Thank you for your response amistre64 but you didn't even answer my question.

Answer 6

Why do they multiply 4x by (dy/dt) and 4y by (dx/dt) ??

Answer 7

I did answer it.

dx/dt = x', dy/dt = y'

how do you work a product rule: what is the derivative of uv?

I demonstrated it, and then took the derivative of (xy)

what about this does not set well with you?

Answer 8

4xy -3x+5y = -40 ; take the derivative with respect to t

D[4xy] -D[3x] +D[5y] = D[-40] or
\[\frac {d(4xy)}{dt}+\frac {d(-3x)}{dt}+\frac {d(5y)}{dt}=\frac {d(-40)}{dt}\]

constants pull out

4D[xy] -3D[x] +5D[y] = D[-40] or
\[4\frac {d(xy)}{dt}-3\frac {d(x)}{dt}+5\frac {d(y)}{dt}=-40\frac {d(1)}{dt}\]

the last 3 terms are self explantory

4D[xy] -3x' +5y' = 0 or
\[4\frac {d(xy)}{dt}-3\frac {dx}{dt}+5\frac {dy}{dt}=-40(0)~\text{or}\]
\[4\frac {d(xy)}{dt}-3x'+5'y=0\]

so the question remains .. what is the derivative of (xy) with respect to t?

how do you work a product rule?