Find the derivative dy/dx = y' using Implicit Differentiation:
x^2 - y^2 = 16

Question

Find the derivative dy/dx = y' using Implicit Differentiation: 
x^2 - y^2 = 16

turingtest · Answer

why don't you try to take the derivative and show me what you get so I can tell you where you go wrong

anonymous · Answer

Is this the answer: 2x-2yy'=0

turingtest · Answer

yes, great!
but we are not done because we have not found dy/dx
so now solve for y'

anonymous · Answer

yeah that's the part where I don't know.

turingtest · Answer

it's just algebra, perhaps the notation is confusing you
what if it was 
2x-2yd=0
and I asked you to solve for d ?
could you do that?

anonymous · Answer

y' = 2x/2y? So these problems have two steps?

amistre64 · Answer

what do you mean by "two steps"?

turingtest · Answer

or three usually
now we need to plug in for y

turingtest · Answer

...not necessarily, but often they want you to get everything in terms of x

-amistre: I think he is counting
1)taking derivative
2)solving for y'

amistre64 · Answer

subbing in for y would have made this more explicit to begin with if you could solve for y ;)

turingtest · Answer

...but it does say to differentiate implicitly

-so the next step would be
3)sub in for y$$x^2-y^2=16\implies y=\sqrt{x^2-16}$$

anonymous · Answer

so put the y'=2x/2y into where?

anonymous · Answer

i dont no

turingtest · Answer

first of all, you can't neglect basic algebra
we found that$$y'=\frac{\cancel2^1x}{\cancel2y}=\frac xy$$do try to remember thigns like the fact that the 2's canceland we have from above that$$x^2-y^2=16\implies y=\sqrt{x^2-16}$$so plugging that expression for y' we get$$y'=\frac xy={x\over\sqrt{x^2-16}}$$