# Differentiability of a function of two variables

• I
I have been studying multivariable calculus but I can't quite think visually how a function will be differentiable at a point.

How can a function be differentiable if its partial derivatives are not continuous?

member 587159
This can even fail in the 1 dimensional case.

Consider ##f(t) = t^2 \sin(1/t)## when ##t \neq 0## and ##f(0) =0##. This function is differentiable on ##\mathbb{R}## but its derivative isn't continuous.

Why would the result hold in the higher derivative case? (Similar counterexamples are possible)

Got it but directing to my first doubt how can differentiability be defined for a function of two variables,like,what is the basic condition for such a function to be differentiable?

hilbert2
Gold Member
For a real-valued function of one real variable, the differentiability means that the difference ##f(x+\Delta x) - f(x)## can be arbitrarily well approximated by a differential ##f'(x)\Delta x##, and the error of this approximation decreases as fast as ##(\Delta x)^2## when the ##\Delta x## is approaching zero. The real number ##f'(x)## is the derivative.

For a real-valued function of two variables, the equivalent definition is that the difference ##f(\mathbf{x}+\mathbf{\Delta x}) - f(\mathbf{x})## can be similarly approximated with a dot product ##\mathbf{f'(x)}\cdot\mathbf{\Delta x}##. Here the derivative ##\mathbf{f'(x)}## is now a two-component vector and it is the same as the gradient of the function ##f(\mathbf{x})##.

If the function takes two arguments and returns a two-component vector, the derivative ##\mathbf{f'(x)}## is a ##2\times 2## matrix and the differential is ##\mathbf{f'(x)}\mathbf{\Delta x}##, where there's a matrix-vector multiplication instead of a dot product.

fresh_42
Mentor
2021 Award
Got it but directing to my first doubt how can differentiability be defined for a function of two variables,like,what is the basic condition for such a function to be differentiable?
Maybe the beginning of https://www.physicsforums.com/insights/the-pantheon-of-derivatives-i/ can help you here. As @hilbert2 has said: differentiability is only the possibilty to approximate a function locally by a linear function. The question, whether these many local events result in a continuous dependency of the location is a completely different one. The most common mistake comes from the notation: ##f\,'(x)##. It is wrong. It should better be ##f\,'(a)##, since this is what differentiation does: it gives a slope at a certain point. Differentiability now means, that this can be done at all points ##x=a##, such that we get a new function ##a \longmapsto f\,'(a)## which people write ##x \longmapsto f\,'(x)## and since people are lazy, abbreviate it by ##f\,'(x)##. The result is, that the dependency of the location, at which ##x \longmapsto f(x)## has been approximated by a linear function via the calculation ##\left.\dfrac{d}{dx}\right|_{x=a}f(x)## is completely lost.

So what differentiablity means is that those linear approximations exist, at a point or everywhere.
Now whether this differentiabilty depends continuously or differentiable from said location is another step.

hilbert2
Gold Member
Yes, I should have emphasized that the point ##x## where the derivative is evaluated is kept constant and only the ##\Delta x## is varied.

Stephen Tashi