Differentiation of functions of several variables is about finding the derivatives of those functions defined on two or more variables. A real-valued function of n-variables is a function f: D→R, where D is an open subset of Rⁿ. Now for any (x₁, x₂, …, x_n) in D, f(x₁, x₂, …, x_n) is a real number. For example, the volume function of a right-circular cone f(x, y) = V ∈ R where

f(x, y) = ⅓(𝜋 x² y); x is the radius of the cone, and y is the height of the cone. Hence, f is a function of two variables.

Likewise, a multivariable function of n-variables is a function f: D → R^m, where D is a subset of Rⁿ. For any (x₁, x₂, …, x_n) in D, f(x₁, x₂, …, x_n) ∈ Rⁿ. For example, if any particle is in a rotational motion such that it is at a distance r from the origin, then at any instant t, the position of the particle is given by f(t) = (r cos (t), r sin (t)). This is a function from R to R².

Now we shall discuss the differentiability of such types of functions.

Differentiability of Functions of Several Variables

Let E be an open set in Rⁿ, and f be a function that maps E into R^m. For (x₁, x₂, …, x_n) = x ∈ E, f(x) ∈ R^m. Suppose there exists a linear transformation A: Rⁿ → R^m such that for any h ∈ Rⁿ

\(\begin{array}{l}\displaystyle \lim_{\textbf{h} \to 0}\frac{|f(\textbf{x}+\textbf{h})-f(\textbf{x})-A\textbf{h})|}{||\textbf{h}||}=0\;\;….(1)\end{array} \)

Then we say that f is differentiable at x, and we write f’(x) = A.

If f is differentiable at every x ∈ E, then f is differentiable in E.

Note that h ∈ Rⁿ, since E is open and x ∈ E, we have x + h ∈ E if ||h|| is small enough. Moreover, f(x) and f(x + h) both are in R^m, also, since A ∈ L(Rⁿ, R^m), Ah ∈ R^m. It follows that

f(x + h) – f(x) – Ah ∈ R^m

Therefore, in (1) the numerator, there is R^m– norm of f(x + h) – f(x) – Ah, and in the denominator, there is Rⁿ– norm of h.

Some Important Results

Unique Existence of Derivative: Let E be an open set in Rⁿ, and f be a function that maps E into R^m. For (x₁, x₂, …, x_n) = x ∈ E if

\(\begin{array}{l}\displaystyle \lim_{\textbf{h} \to 0}\frac{|f(\textbf{x}+\textbf{h})-f(\textbf{x})-A\textbf{h})|}{||\textbf{h}||}=0\end{array} \)

such that A = A₁ and A = A₂. Then A₁ = A₂. That is, the derivative of f exists uniquely.

Let E be an open set in Rⁿ, and f be a function that maps E into R^m. For (x₁, x₂, …, x_n) = x ∈ E if f is differentiable at x. Then f is continuous at x.
Suppose a function of several variables f: E → R^mwhere E is an open subset of Rⁿ, that is f defined on n variables. For (x₁, x₂, …, x_n) = x ∈ E if f is differentiable at x then the partial derivates f_x1, f_x2, …, f_xn exists and the derivative f’(x) = (𝛼₁, 𝛼₂, …, 𝛼_m).

\(\begin{array}{l}\alpha_{i}=f_{x_{i}}=\frac{\partial f}{\partial x_{i}}\;\;\;\; 1\leq i\leq n\end{array} \)

Generalised Version of Chain Rule: Let E be an open set in Rⁿ, and f be a function that maps E into R^m. Let f be differentiable at x_o ∈ E, g maps an open set containing f(E) into R^k and g be differentiable at f(x_o). Then the mapping F: E → R^k is defined by F(x) = g(f(x)) is differentiable at x_o and F’(x_o) = g’(f(x_o)) f’(x_o).
Existence of Derivatives: Let E be an open set in Rⁿ, and f be a function that maps E into R^m. If all partial derivatives of f exist in the neighbourhood of point x_oin E and f is continuous at x_o. Then f is differentiable at x_o_.

Solved Examples

Example 1:

Let a function f be defined as

\(\begin{array}{l}f(x,y)= \left\{\begin{matrix}xy\left(\frac{x^{2}-y^{2}}{x^{2}+y^{2}}\right) \;\;\;\; if\;(x,y)\neq(0,0) \\0\;\;\;\; \;\;\;\;\;\;\;if \;(x,y)=(0,0)\end{matrix}\right.\end{array} \)

Prove that f is differentiable at (0, 0).

Solution:

To verify the differentiability of f, at (0, 0), let us take A = (f_x, f_y) at (0, 0)

Both f_x and f_y at (0, 0) is 0. Therefore A = (0, 0).

Let x = (x, y) and h = (h, k), we shall prove that

\(\begin{array}{l}\frac{|f(\textbf{x}+\textbf{h})-f(\textbf{x})-A\textbf{h})|}{||\textbf{h}||}\rightarrow 0 \:\:\:\: as \:\:\:\:\textbf{h}\rightarrow 0\end{array} \)

Now, f(0 + h) = f((0, 0), (h, k)) = f(0 + h, 0 + k) = f(h, k) = hk[(h² – k²)/(h² + k²)]

f(0) = f((0, 0)) = 0

And Ah = (0, 0)(h, k) = 0

\(\begin{array}{l}\displaystyle \lim_{\textbf{h} \to 0}\frac{|f(\textbf{x}+\textbf{h})-f(\textbf{x})-A\textbf{h})|}{||\textbf{h}||}= \displaystyle \lim_{(h,k) \to 0}\frac{|hk\left(\frac{h^{2}-k^{2}}{h^{2}+k^{2}}\right)-0-0|}{\sqrt{h^{2}+k^{2}}}\end{array} \)

\(\begin{array}{l}=\displaystyle \lim_{(h,k) \to 0}hk \;\;\displaystyle \lim_{(h,k) \to 0}\frac{h^{2}-k^{2}}{(h^{2}+k^{2})^{\frac{3}{2}}}=0 \end{array} \)

Hence, f is differentiable at 0.

Example 2:

Let a function f be defined as

\(\begin{array}{l}f(x,y)=\left\{\begin{matrix} \frac{2xy}{x^2+y^2}& at\:\:(x,y)\neq (0,0) \\0 & at\:\:(x,y)= (0,0) \\\end{matrix}\right\end{array} \)

Prove that f is differentiable at (0, 0).

Solution:

Given,

\(\begin{array}{l}f(x,y)=\left\{\begin{matrix} \frac{2xy}{x^2+y^2}& at\:\:(x,y)\neq (0,0) \\0 & at\:\:(x,y)= (0,0) \\\end{matrix}\right\end{array} \)

We see that f(0, 0) = 0 is not continuous at (0, 0). However the partial derivatives exists at (0, 0).

Therefore, f is not differentiable at (0, 0).

Frequently Asked Questions

How to find the derivative of a multivariate function at a given point?

Suppose the multivariate function is differentiable at a given point. Then its derivative is equal to the partial derivative of the function with respect to each variable.

What is meant by a function of several variables?

A function f which maps from Rⁿto R^m, that is, f: Rⁿ → R^m, is known as a function of several variables. If x = (x₁, x₂, …, x_n) ∈ Rⁿ then f(x) ∈ R^m.

How to determine the differentiability of a function of several variables?

If f is a function of several variables, f: Rⁿ → R^m, f is said to be differentiable at x_o∈ Rⁿif the quantity [|f(x_o + h) – f(x_o) – Ah|/||h||] → 0 as h → 0, where h ∈ Rⁿsuch that || h || is very small and f’(x_o) = A.

What is the necessary condition for a function of several variables to be continuous at a given point?

Let E be an open set in Rⁿ, and f be a function that maps E into R^m. For (x₁, x₂, …, x_n) = x ∈ E if f is differentiable at x. Then f is continuous at x.