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:
?
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:
?
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'
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!