Implicit Differentiation Example: Finding Solutions for f(x,y) = 0

[Rasalhague]

Spivak: Calculus on Manifolds, p. 42:

Reconsider the function f:\mathbb{R}^2 \rightarrow \mathbb{R} defined by f(x,y) = x^2 + y^2 - 1, we note that the two possible functions satisfying f(x,g(x)) = 0 are

g(x) = \sqrt{1-x^2}

and

g(x) = -\sqrt{1-x^2}.

Differentiating f(x,g(x)) = 0 gives

D_1(x,g(x))+D_2(x,g(x)) \cdot g'(x) = 0

or

2x +2g(x) \cdot g'(x) = 0,

g'(x) = -x/g(x),

which is indeed the case for either

g(x) = \sqrt{1-x^2}

or

g(x) = -\sqrt{1-x^2}.

Assuming "differentiating f(x,g(x)) = 0" means differentiating f \circ h, where

h : (-1,1) \rightarrow \mathbb{R}^2 \; | \; x \mapsto (I(x),g(x)), I(x) = x

and setting the result identically equal to zero, we have

2x+2g(x) \cdot g'(x) = 0,

as above. Substituting

g(x) = \sqrt{1-x^2},

I get

=2x-\frac{2 \sqrt{1-x^2} \cdot 2x}{\sqrt{1-x^2}}

=-2x = 0.

Or, if I substitute the negative square root, 6x = 0. Therefore, either way, x = 0. But hang on! By definition, g maps (-1,1) to the reals, its graph forming a semicircle; how can this other equation be telling us that x=0?

 

[AlephZero]

g(x) = (1 - x^2)^{1/2}

g'(x) = (-2x)(1/2) (1-x^2)^{-1/2}

You seem to be have lost a factor of 1/2.

 

[Rasalhague]

Ah yes, I see. Thanks, AlephZero!

When g(x)=(1-x²)^1/2, we have 2x - 2x = 0, which is consistent and doesn't set any restriction on x. Likewise when g(x)=-(1-x²)^1/2, since the minus from g(x) "cancels" the minus from g'(x), giving the same result: 2x - 2x = 0.