Evaluate dy/dt for each of the following

[staples82]

Homework Statement

Assume x and y are functions of t. Evaluate dy/dt for each of the following.

xy-5x+2y^3=-70, dx/dt=-5 x=2 y=-3

Homework Equations

n/a

The Attempt at a Solution

I used x' and y' for dx/dt and dy/dt

I found the derivative of the equation, using the product rule for xy

(x)(yy')+(y)(xx')-5+6y^2=0

From here, my problem was figuring out how to simplify the equation...
(x)(yy')=5-6y^2-yxx'
y'=

The answer was -5/7...I'm pretty sure I messed up on solving the equation for 0 and y'

 

[dynamicsolo]

Homework Statement

Assume x and y are functions of t. Evaluate dy/dt for each of the following.

xy-5x+2y^3=-70, dx/dt=-5 x=2 y=-3

...

I found the derivative of the equation, using the product rule for xy

(x)(yy')+(y)(xx')-5+6y^2=0

This is not differentiated correctly with respect to the third variable. When dealing with the product rule for the xy term, think of x and y as if both were functions of t. This would have to be

d/dt (xy) = (dx/dt)·y + x·(dy/dt),

or x'y + xy' , as you have been writing it. By the same token, the next term would have the derivative -5x' .

Keep in mind that you are not differentiating with respect to x, as you were in your other implicit differentiation problems, but with respect to a variable whichi is not directly (we say "explicitly") represented in the original equation.

 

[staples82]

So I messed up my product rule of xy then?

 

[Defennder]

Yeah I believe so. What is the derivative of xy with respect to t?

 

[staples82]

wouldn't it be: x(dy/dt)+y(dx/dt) for the derivative of xy

 

[dynamicsolo]

wouldn't it be: x(dy/dt)+y(dx/dt) for the derivative of xy

That's correct (and is mentioned in post #2). You need to differentiate the remaining terms on the left-hand side of the original equation in the same fashion. So d/dt (5x) and d/dt (2y^3) are not simply 5 and 6y^2, but...?