# Math Help - Differentiating function that contains y and x?

1. ## Differentiating function that contains y and x?

hey guys can you help me with this problem.

Given that x^2 + x – y^2 = 0

a) find dy/dx.

is it this:
$\frac{2x+1}{2y} = \frac{dy}{dx}$

b) Find dy/dt at x = 5 if dx/dt = 2/3

what do I do here?

is it:
$\frac{dy}{dt} =\frac{dy}{dx} *\frac{dx}{dt}$
$\frac{dy}{dt} = \frac{2(5)+1}{2y} = \frac{dy}{dx} * \frac{2}{3}$?
$\frac{11}{3y} = \frac{dy}{dt}$

2. Originally Posted by RezMan
hey guys can you help me with this problem.

Given that x^2 + x – y^2 = 0

a) find dy/dx.

is it this:
$\frac{2x+1}{2y} = \frac{dy}{dx}$

b) Find dy/dt at x = 5 if dx/dt = 2/3

what do I do here?

is it:
$\frac{dy}{dt} =\frac{dy}{dx} *\frac{dx}{dt}$
$\frac{dy}{dt} = \frac{2(5)+1}{2y} = \frac{dy}{dx} * \frac{2}{3}$?
$\frac{11}{3y} = \frac{dy}{dt}$
It's called "implicit differentiation".
Basically, you can navigate through this if you always use the chain rule, and remember that dx/dx = 1 (= x', when differentiating w.r.t. x)

So for
x^2 + x – y^2 = 0
(x^2)' + (x)' - (y^2)' = 0
2x*x' + 1*x' -2y*y' = 0
2x + 1 - 2y*y' = 0
2x + 1 = 2y*y'
(2x + 1)/2y = y'

3. Thankyou.

so what do I do for the second part?

4. It's like the first way, but x' = dx/dt (NOT 1!) and y' = dy/dt
(First, you'll want to put x = 5 into the original equation and solve for y)

x^2 + x – y^2 = 0
(x^2)' + (x)' - (y^2)' = 0
2x*x' + 1*x' -2y*y' = 0
Then use the given values of x' = dx/dt = 2/3, x = 5, y = (whatever you found it to be above) and solve for y' = dy/dt

And I'm sticking with "whatsoever ye do, whether in word or deed..."

edit...
Oh, yeah, I guess you could work from your result in part (a). That would make some sense!

5. does this frankenstien monster look like it:

$2x(\frac{2}{3})+1(\frac{2}{3})-2\sqrt{30}(\frac{2x+1}{2y})$

yes?

6. so would you say this is correct 11/3y=dy/dt

7. Originally Posted by RezMan
so would you say this is correct 11/3y=dy/dt
Where y = +/-√30
If you're going this route, you can forego the steps I suggested in post #4.
If you needed to find dy/dt from scratch, you would take the steps in post #4!

8. okay cool. I think I've been able to put this to rest lol