# excercise about implicit functions

Let $f(x,y)=\cos(x^2)+2xy+\sin(y^2)-4x-1+y$.

Show that there exists an environment around zero so that $f(x,y)=0 \iff y=g(x)$ for a unique function $g$. Further show that $g$ is two times continuous differentiable in this environment and calculate $g'(0)$ and $g''(0)$.

$f$ is continuous differentiable, which can be seen with help of the differential matrix:

$$\mathrm{D}f(x,y)=\left(-2x\sin(x^2)+2y-4 \quad ,\quad 2x+2y\cos(y^2)-1\right)$$

So $f$ is partial differentiable and both partial derivatives are composed of continuous differentiable functions. So $f$ is continuous differentiable.

It is $f(0,0)=0$ und $D_2f(0,0)=-4$ so $D_2f(0,0)$ is invertible. It exists a unique function $g$ with $y=g(x)$ and

$$g^\prime(x)=Dg(x)=-\frac{\partial f(x,y)}{\partial y}^{-1}\frac{\partial f(x,y)}{\partial x}$$

It follows $g^\prime(x)=\frac{2x\sin(x^2)-2y+4}{2x+2y\cos(y^2)+1}$ which is continuous and differentiable and $g^\prime(0)=\frac{2y+44}{2y\cos(y^2)+1}=\frac{2g(0)+4}{2g(0)\cos((g(0))^2)+1}=4$

Let now $y=g(x)$ and differentiate $g^\prime(x)$

Is this ok so far? But now differentiating $g'(x)$ becomes a really exhaustive task. Is there a simpler way to do this than applying quotient rule and fill several pages?

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Dear J., in mathematical English we usually say "neighbourhood" rather than "environment" . –  Georges Elencwajg Jul 2 '12 at 17:56

Just continue the same reasoning that leads to the formula for $g'(x)$. Namely, we have $$0=\frac{d}{dx}f(x,y)=D_1f+\frac{dy}{dx}\cdot D_2f$$ Now differentiate again: $$0=D_{11}f+2\frac{dy}{dx}\cdot D_{12}f+\left(\frac{dy}{dx}\right)^2\cdot D_{22}f+\frac{d^2y}{dx^2}\cdot D_2f$$ and you can solve for $g''(x)=\frac{d^2y}{dx^2}$ from here.