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.