hey guys can you help me with this problem.

Given that x^2 + x – y^2 = 0

a) find dy/dx.

is it this:

b) Finddy/dtat x = 5 ifdx/dt= 2/3

what do I do here?

is it:

?

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- May 8th 2011, 05:54 PMRezManDifferentiating 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:

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

what do I do here?

is it:

?

- May 8th 2011, 06:30 PMTheChaz
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' - May 8th 2011, 06:57 PMRezMan
Thankyou.

so what do I do for the second part? - May 8th 2011, 07:07 PMTheChaz
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! - May 8th 2011, 07:26 PMRezMan
does this frankenstien monster look like it:

yes? - May 8th 2011, 07:50 PMRezMan
so would you say this is correct 11/3y=dy/dt

- May 8th 2011, 07:58 PMTheChaz
- May 8th 2011, 08:06 PMRezMan
okay cool. I think I've been able to put this to rest lol