Partial derivatives play a central role in multivariable calculus. As the derivative of single variable functions, partial derivatives can be interpreted as rates of change and slopes of tangents while studying functions of several variables. Partial derivatives are also used in linear or tangent plane approximation of surfaces represented by functions of two variables. In the study of vector calculus, partial derivatives form the component functions in gradient vectors.
Let us look how partial derivatives are defined and learn how to find them.

A Partial derivative in a function of several variables is the rate of change of the function, by allowing one variable to vary while freezing all the other variables to constants.

Let f(x, y) be a function of two variable x and y. Suppose y is fixed as a constant y = b and only the variable x is allowed to change. Then we are really considering a function of one variable x and we can denote this as g(x) = f(x, b). Hence the partial derivative of f(x, y) with respect to x is indeed g'(x) and denoted by fx.

Similarly the partial derivative of f(x, y) with respect to y is denoted by fy and it is equal to h'(y) where h(y) = f(y, b).
The commonly used Leibniz notation for partial derivatives are:

$\frac{\partial f}{\partial x}$ representing the partial derivative of f with respect to x and

$\frac{\partial f}{\partial y}$  representing the partial derivative of f with respect to y.

Limit Definition of Partial derivatives
The partial derivative of f(x, y) with respect to x at the point (a, b) is

$\frac{\partial f}{\partial x}_{(a,b)}$ = $\lim_{h->0}\frac{f(a+h,b)-f(a,b)}{h}$

The Partial derivative of f(x, y) with respect to y at the point (a, b) is
$\frac{\partial f}{\partial y}_{(a,b)}$ = $\lim_{h->0}\frac{f(a,b+h)-f(a,b)}{h}$

Geometrically a partial derivative of a function of two variables can be interpreted as the slopes of the tangents to the traces of the surface represented by the function.
If S is the surface represented by the function
z = f(x, y), then C1 and C2 are the traces cut off
by the planes y = b and x = a at the point
P(a, b, c).  The partial derivatives fx and fy represent
slopes of the tangent lines (T1 and T2) to the curves
C1 and C2.

 Partial Derivative Interpretation

The product, quotient and chain rule for partial derivatives are applied in a manner similar to these methods in ordinary differentiation. Let us work out an example to show this.

Solved Example

Question: Find $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$  for the function f(x,y) = e(x+y)xy
Solution:
 
First to find $\frac{\delta f}{\delta x}$ we treat y as a constant an differentiate using product rule.
f(x,y) = yey (exx)                                                    Constant terms are pulled out.
$\frac{\partial f}{\partial x}$  = yey (ex + ex x)                                             Function is x is differentiated using product rule
                                         = yey ex (1 + x)
                                         = y e(x+y) ( 1 + x)
Similarly it can be shown that
$\frac{\partial f}{\partial y}$ = xex+y (1 + y)

 

Partial derivatives $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ are also functions of x,y. Hence the partial derivation process can be continued on these functions to get the second order partial derivatives like fxx , fxy, fyy and fyx.
(fx)x = fxx = $\frac{\partial }{\partial x}(\frac{\partial f}{\partial x})$ = $\frac{\partial ^{2}f}{\partial x^{2}}$

(fx)y = fxy = $\frac{\partial }{\partial y}(\frac{\partial f}{\partial x})$ = $\frac{\partial ^{2}f}{\partial y\partial x}$

(fy)x = fyx = $\frac{\partial }{\partial x}(\frac{\partial f}{\partial y})$ = $\frac{\partial ^{2}f}{\partial x\partial y}$

(fy)y = fyy = $\frac{\partial }{\partial y}(\frac{\partial f}{\partial y})$ = $\frac{\partial ^{2}f}{\partial y^{2}}$

Notation fxy or ($\frac{\partial ^{2}f}{\partial y\partial x}$) means that we first differentiate f with respect to x then with respect to y.
Clairaut's Theorem: Suppose f is defined on a Disk contains the point (a, b). If the functions fxy and fyx are both continuous on D then,
fxy (a,b) = fyx(a,b)

The chain rule we apply for finding the ordinary derivative can be applied partial derivatives when the function is directly defined as a function of two or more variables. But in turn if these variables are defined as functions of one or more other variables, then the chain rule is applied as it is applied in the context of a function defined using parameters.
Case 1:
When the variables x and y are functions of one variable t
Let z = f(x, y) and x = g(t) and y = h(t). This means z is indirectly a function of t and its derivative with respect to t can be found at a point if x and y are differentiable with respect to t at that point.
$\frac{dz}{dt}= \frac{\partial z}{\partial x}.\frac{dx}{dt}+\frac{\partial z}{\partial y}.\frac{dy}{dt}$

Case 2:
When the variables x and y are functions of two variables u and v
Let z = f(x, y) and x = g(u, v) and y = h(u, v). z is thus indirectly the function of the variables u and v. Hence the partial derivatives of z with respect to u and v can be found using the formulas,
$\frac{\partial z}{\partial u}= \frac{\partial z}{\partial x}.\frac{\partial x}{\partial u}+\frac{\partial z}{\partial y}.\frac{\partial y}{\partial u}$
and
$\frac{\partial z}{\partial v}= \frac{\partial z}{\partial x}.\frac{\partial x}{\partial v}+\frac{\partial z}{\partial y}.\frac{\partial y}{\partial v}$

Chain rule General version:
Extending the above formulas the general version of the chain rule for finding the partial derivatives of a function of n variables, which in turn are functions of m other variables, can be stated as follows:
$\frac{\partial z}{\partial t_{i}}$ = $\frac{\partial z}{\partial x_{1}}.\frac{\partial x_{1}}{\partial t_{i}}+\frac{\partial z}{\partial x_{2}}.\frac{\partial x_{2}}{\partial t_{i}}+..............+\frac{\partial z}{\partial x_{n}}.\frac{\partial x_{n}}{\partial t_{i}}$
where i can assume values 1, 2, 3....m.

Solved Examples

Question 1: f(x, y) = 3x2 + x3y + 4y2, find the second partial derivatives $\frac{\partial ^{2}f}{\partial x^{2}}$$\frac{\partial ^{2}f}{\partial    y^{2}}$, $\frac{\partial ^{2}f}{\partial y\partial x}$ and $\frac{\partial ^{2}f}{\partial x\partial y}$. Also verify Clairaut's theorem.
Solution:
 
First let us compute first order partial derivatives.
$\frac{\partial f}{\partial x}$ = 6x + 3x2y                        y is treated as a constant in f(x, y).

$\frac{\partial f}{\partial y}$ = x3 + 8y                           x is treated as a constant in f(x, y).

Now let us find the second order partial derivatives by differentiating again partially the first partial derivatives.

$\frac{\partial ^{2}f}{\partial x^{2}}$ = $\frac{\partial }{\partial x}(\frac{\partial f}{\partial x})$  = 6 + 6xy

$\frac{\partial ^{2}f}{\partial y^{2}}$ = $\frac{\partial }{\partial y}(\frac{\partial f}{\partial y})$ = 8

$\frac{\partial ^{2}f}{\partial y\partial x}$ = $\frac{\partial }{\partial y}(\frac{\partial f}{\partial x})$
         = $\frac{\partial }{\partial y}$
(6x + 3x2y) = 3x2.

$\frac{\partial ^{2}f}{\partial x\partial y}$ = $\frac{\partial }{\partial y}(\frac{\partial f}{\partial y})$
         =$\frac{\partial }{\partial x}$
( x3 + 8y ) = 3x2.

f(x, y) being a polynomial in x and y is defined everywhere in R2 and both the second order derivatives $\frac{\partial ^{2}f}{\partial y\partial x}$  and $\frac{\partial ^{2}f}{\partial x\partial y}$ are continuous in R2. The conditions for Clairaut's theorem hold it is also shown
$\frac{\partial ^{2}f}{\partial y\partial x}$ = $\frac{\partial ^{2}f}{\partial x\partial y}$  = 3x2.

 

Question 2: Compute the second order partial derivatives of f(x, y) = sin (2x + 3y)
Solution:
 
fx = 2cos (2x + 3y)
   fy  = 3cos(2x + 3y)
   fxx = -4sin(2x + 3y)
   fyy = -9sin(2x +3y)
   fxy = -6sin(2x + 3y)
   fyx = -6sin(2x + 3y)