Partial differentiation is used to differentiate mathematical functions having more than one variable in them. In ordinary differentiation, we find derivative with respect to one variable only, as function contains only one variable. So partial differentiation is more general than ordinary differentiation. Partial differentiation is used for finding maxima and minima in optimization problems. Let us discuss it in details.

**Browse more Topics under Calculus**

- Application of Marginal Cost And Marginal Revenue
- Introduction to Integral Calculus
- Methods of Integration

**Partial Differentiation**

It’s another name is Partial Derivative. It is a derivative where we hold some independent variable as constant and find derivative with respect to another independent variable.

For example, suppose we have an equation of a curve with X and Y coordinates in it as 2 independent variables. To find the slope in the direction of X, while keeping Y fixed, we will find the partial derivative. Similarly, we can find the slope in the direction of Y (keeping X as Fixed).

Partial differentiation builds with the use of concepts of ordinary differentiation. So we should be familiar with the methods of doing ordinary first-order differentiation.

Obviously, for a function of one variable, its partial derivative is the same as the ordinary derivative. Constants, when added to functions, will differentiate to 0 but constants multiplying the functions, are retained and so still multiply the derivative.

### Mathematical Representation

Here f’x to mean “the partial derivative with respect to x”. Its another very common notation is to use a backward d (∂) like this: ∂f∂x = 2x

This is the same as: f’x = 2x ∂ is also called “del” or “dee” or “curly dee”

__Example with Explanation__

__Example with Explanation__

Take a function of one variable x:

f(x) = x^{2}

It’s derivative using power rule:

First order derivative :: f’(x) = 2x

Now take a function of two variables x and y:

f(x,y) = x^{2} + y^{3}

To find its partial derivative with respect to x we consider y as a constant:

Partial derivative wrt X :: f’_{x} = 2x + 0

= 2x

Now, to find the partial derivative with respect to y, we consider x as a constant:

Partial derivative wrt Y :: f’_{y} = 0 + 3y^{2}

= 3y^{2}

For more than two variables we will consider all other variables as if they are constants.

Source: freepik.com

__Practical Implication__

__Practical Implication__

Consider a Cylindrical object. The equation to find volume is:

V = **π** r^{2} h

Also, We can write that in multi-variable form as f(r,h) = **π** r^{2} h

For the partial derivative with respect to r we hold h constant, and r changes: f’_{r} = **π** (2r) h = 2**π**rh

Here derivative of r^{2} with respect to r is 2r, and π is a constant and we assume h as constant. It says that, as only the radius changes (by the negligible amount), the volume changes by 2**π**rh. It is like that we add another skin with a circle’s circumference 2**π**r and a height of h.

For the partial derivative with respect to h, we hold r as a constant:

f’_{h} = **π** r^{2 }(1)

= **π**r^{2}

**π** and r^{2} both are constants, and the derivative of h with respect to h is 1.

It says that as only the height changes (by the negligible amount, the volume changes by **π**r^{2}.

It is like we add the thinnest disk with negligible height on top with a circle’s area of **π**r^{2}.

**Solved Example on Partial Differentiation**

**Question-1:** Find the partial derivative of the following function (in x and y) with respect to x and y separately. f(x,y) = 2x^{2} + 4xy

Answer: With respect to X : f’_{x} = 4x + 4y

With respect to Y : f’_{y} = 0 + 4x = 4x

**Question-2 :** Find the partial derivatives of function g given as:

g(x,y,z) = x^{4} − 3xyz.

Answer: Function is g(x,y,z) = x^{4} − 3xyz

*Wrt x :: ∂g***∂x** = 4x^{3} − 3yz

*Wrt y :: ∂g***∂y** = −3xz

*Therefore Wrt z :: ∂g***∂z** = −3xy