Overview
When we find the area of a triangle then it depends on base and height, hence we can say that area of the triangle is the function of base and height. In other words, when we have a mathematical function having more than one variables, we use partial differentiation.
Symbolically and mathematically the relationship is defined as below
z is called a function of two variables x and y if z has one definite value for every pair of values of x and y.
it is written as
z = f (x, y)
The variable x and y are called independent variables while z is called the dependent variable.
Note- the function z = f (x, y) represents a surface.
Partial derivatives
First order partial differentiation–
Let f(x , y) be a function of two variables. Then the partial derivative of this function with respect to x can be written as and defined as follows:
Now the partial derivative of f with respect to f can be written as and defined as follows:
Note: a. while calculating partial derivatives treat all independent variables, other than the variable with respect to which we are differentiating , as constant.
b. we apply all differentiation rules.
Higher order partial differentiation-
Let f(x , y) be a function of two variables. Then its second-order partial derivatives, third order partial derivatives and so on are referred as higher order partial derivatives.
These are second order four partial derivatives:
b and c are known as mixed partial derivatives.
Similarly we can find the other higher order derivatives.
Solved examples
Example-1: Calculate for the following function
f(x , y) = 3x³-5y²+2xy-8x+4y-20
Sol. To calculate the variable y as a constant, then differentiate f(x,y) with respect to x by using differentiation rules,
Similarly partial derivative of f(x,y) with respect to y is:
Example-2: Calculate for the following function
f( x, y) = sin(y²x + 5x – 8)
Sol. To calculate the variable y as a constant, then differentiate f(x,y) with respect to x by using differentiation rules,
Similarly the partial derivative of f(x,y) with respect to y is,
[sin(y²x + 5x – 8)]
= cos(y²x + 5x – 8) (y²x + 5x – 8)
= 2xycos(y²x + 5x – 8)
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