Differentiability

In calculus, a function $f$ is differentiable at $x_0\in I$ if the following limit, the definition of the derivative, exists:

$$\underset{x\to x_0}{\lim }\frac{f(x)-f(x_0)}{x-x_0}$$

If $f$ is differentiable at each point $x\in I$, then $f$ is differentiable. If $f$ is differentiable and $f^{\prime }$ is continuous, then $f$ is continuously differentiable, or of class $C^1$.

In multivariable calculus, we define differentiability using this definition:

Multivariable differentiability

Let $z=f(x,y)$. $f$ is differentiable at $(a,b)$ if $f_x$ and $f_y$ exist and the following are equal:

$$\Delta z=f(a+\Delta x,b+\Delta y)-f(a,b)$$ $$\Delta z=f_x(a,b)\Delta x+f_y(a,b)\Delta y+{ϵ}_1\Delta x+{ϵ}_2\Delta y$$

where ${ϵ}_1$ and ${ϵ}_2$ are functions that only depend on $\Delta x$ and $\Delta y$ with $({ϵ}_1,{ϵ}_2)\to (0,0)$.

By theorem (differentiability implies continuity), if a function is differentiable at $(a,b)$, then it is continuous at $(a,b)$. Conversely if it is not continuous at the point, then it is not differentiable.

Complex analysis

We say that a complex function $f(z)=u+iv$ is complex differentiable at $z_0$ if the below limit exists:

$$\underset{z\to z_0}{\lim }\frac{f(z)-f(z_0)}{z-z_0}=f^{\prime }(z_0)$$

If $f$ is complex differentiable at $z_0=x_0+i{y}_0$, then $u(x,y)$ and $v(x,y)$ will be differentiable (in the real sense) at $(x_0,y_0)$.

Similar to the real case, if $f$ is differentiable at a point, then it is continuous.

Continuity of partial derivatives

By theorem, if $f$ has partial derivatives $f_x$ and $f_y$ defined on an open set containing $(a,b)$, with both continuous at $(a,b)$, then $f$ is differentiable at $(a,b)$.

By contrapositive argument, if $f(x,y)$ is not differentiable at that point, then at least one of the partial derivatives is not continuous at the point or there is no open set containing that point such that the partial derivatives are defined at all points on that set.

But! The inverse isn’t true. If we first say the partials are not continuous, we can’t conclude anything about the differentiability of $f$.

Other definitions

Multivariable differentiability (formally)

Let $z=f(x,y)$, where $(x,y)\in D$. Suppose that the partial derivatives $f_x$ and $f_y$ exist at a point $(a,b)$. We define the tangent plane that touches the surface at $(a,b,f(a,b)$ as:

$$L_{(a,b)}(x,y)=f(a,b)+f_x(a,b)(x-a)+f_y(a,b)(y-b)$$

$f$ is differentiable at $(a,b)$ if:

$$\underset{(x,y)\to (a,b)}{\lim }\frac{f(x,y)-L}{\Vert (x-a,y-b)\Vert }=0$$

$E_d$ is the vertical distance between a position on the tangent plane and the surface. If either of the partial derivatives $f_x$ and $f_y$ do not exist or the limit is not 0, then $f$ is not differentiable at $(a,b)$. But! The definition is generally difficult to verify. So we use the theorem involving the continuity of partial derivatives.

A key implication of this is that if the tangent plane exists at a point (and the partials are continuous), then the function is differentiable at that point.

Single-variable case

The derivative of a function $f(x)$ is differentiable at $x=a$ provided $f^{\prime }(x)$ exists (i.e., by limit definition, it exists), and the following are equal:

$$\Delta y=f(a+\Delta x)-f(a)$$ $$\Delta y=f^{\prime }(x)\Delta x+{ϵ}_1\Delta x$$

where ${ϵ}_1$ is a function that depends only on $\Delta x$ with ${ϵ}_1\to 0$ as $\Delta x\to 0$.