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

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

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

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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)?
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.

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