1/x+1/y=4, and y(5)=5/19. Find y'(5) by implicit differentiation.

Question

anonymous · Answer

Differentiating implicitly w.r.t. x we get:
$$-1/(x^2) + -1/(y^2)*(dy/dx)=0$$

Rearranging:
$$dy/dx = -y^2/x^2$$

So dy/dx evaulated at x=5 is -y(5)^2/25 = -1/(19^2) = -1/361

anonymous · Answer

Why isn't it -1/(x^2)+y'/(y^2)=0?

anonymous · Answer

OK differentiating the 1/y term we have:
$$(d/dx)1/y = (d/dx)y^{-1} = -1y^{-2}*(dy/dx)$$

as I originally posted.  Your sign is wrong.

anonymous · Answer

What I mean is, why aren't you treating y as a function in this case? Or are you notating that by saying it needs to multiply (dy/dx)?

anonymous · Answer

We are treating it as a function; we use the following rule:
$$(d/dx)f(y(x)) = (d/dy)f(y)*(dy/dx)$$ for an arbitary function f

I suggest you read an introductory text to differentiation if you do not understand.  this *is* implicit differentiation.