Theory

Introduction

There is a relation between the concepts of linear approximation, partial derivative, tangent plane, and differentiability. This relation is contained in a function defined in terms of $f(x,y)$:

Linear approximation function

$$L(x,y)=f(a,b)+f_x(a,b)\cdot (x-a)+f_y(a,b)\cdot (y-b).$$

The linear approximation function includes both partial derivatives $f_x$ and $f_y$. The set of points satisfying $z=L(x,y)$ defined by the above equation, namely the graph of $L(x,y)$, is the tangent plane. Lastly, the function $f(x,y)$ is said to be differentiable when:

$$f(x,y)=L(x,y)+(x-a)\cdot {\epsilon }_1+(y-b)\cdot {\epsilon }_2,$$

with two functions ${\epsilon }_1$ and ${\epsilon }_2$ satisfying ${\epsilon }_1,{\epsilon }_2\to 0$ as $(x,y)\to (a,b)$.

If $f(x,y)$ is not differentiable, then the linear approximation may be a bad approximation.

Linear approximation in 2d

Remember that for a function $f(x)$ of a single variable, we have the linear approximation:

$$L(x)=f(a)+f^{\prime }(a)\cdot (x-a).$$

The tangent line to $f$ at $x=a$ is given by $y=L(x)$, and $f$ is considered differentiable when $f(x)=L(x)+(x-a)\cdot \epsilon$ such that $\epsilon \to 0$ as $x\to a$:

$$\left(\frac{f(x)-f(a)}{x-a}-f^{\prime }(a)\right)=\epsilon \to 0\text{ as }x\to a\text{iff }{f}^{\prime }(x)\text{ exists at }a.$$

Tangent plane via tangent lines

How do we know that the graph of $L(x,y)$ gives the tangent plane?

The tangent vector to the trace $z=f(x,b)$ at $x=a$ is given by ${\frac{d}{dx}(x,b,f(t,b))|}_a=(1,0,f_x(a,b))$. So the tangent line is given by:

$${\mathbf{q}}_{(a,b)}(t)=(a,b,f(a,b))+t(1,0,f_x(a,b))=(a+t,b,f(a,b)+t{f}_x(a,b)).$$

These points satisfy $z=L(x,y)$. The graph $z=L(x,y)$ must be the tangent plane because it is the unique plane such that the tangent lines of the trace curves $f(x,b)$ and $f(a,y)$ lie in this plane.

Taylor series and linear approximation

Linear approximation gives the first two terms of the Taylor series of a function, and the remainder term, captured in $\epsilon$, must limit to zero when the function is differentiable:

$$\begin{array}{rl}f(x)& =f(a)+f^{\prime }(a)(x-a)+\frac{1}{2!}{f}^{\prime \prime }(a)(x-a{)}^2+\dots \\ & =f(a)+f^{\prime }(a)(x-a)+\epsilon \cdot (x-a).\end{array}$$

If $f$ agrees with its Taylor series, then:

$$\epsilon (x)=\frac{1}{2!}{f}^{\prime \prime }(a)(x-a)+\frac{1}{3!}{f}^{(3)}(a)(x-a{)}^2+\frac{1}{4!}{f}^{(4)}(x-a{)}^3+\dots$$

Differentials

Some authors use the mnemonic notation of finite differentials for applications of linear approximation. These are written as $dx$, $dy$, $dz$, $df$, etc., even though they are not infinitesimals. Their usage is according to the following kind of rules:

$$df=\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy.$$

Differentiability

In 2d, a function $f(x)$ is differentiable precisely where its derivative $f^{\prime }(x)$ exists. The situation is not so simple in 3d: a function like $f(x,y)=\frac{2xy(x+y)}{x^2+y^2}$ has both partial derivatives at $(x,y)=(0,0)$, but it is not differentiable there! On the other hand, one theorem does help:

Theorem

If a function $f(x,y)$ is given and both partials $f_x$ and $f_y$ exist and are continuous at a point $(a,b)$, then $f$ is differentiable at $(a,b)$.

Differentiability (part 2)

Differentiable, but partials aren’t continuous: $r^2\sin (\frac{1}{r})$

The function $f(r,\theta )=r^2\sin (\frac{1}{r})$ is differentiable at $(0,0)$ since it is approximated by its tangent plane at $(0,0)$. The partials are not continuous there! You may check this by computing $f_x$ and $f_y$. You can also compute $f_r=2r\sin (1/r)-\cos (1/r)$, which is not continuous at $r=0$.