Subject: Calculus

# Partial Derivative

So far, we have only discussed functions with single variable or parameter. However, in the real world, physical quantities and phenomenon depends on two or more variables. From here on, we will be extending the basic ideas of differential calculus to these kinds of function.

To better understand this concept, let us consider things that we know so well. For example, the volume V of a cylinder is dependent on its radius r and height h. We can express it as a function of two variables and write volume as V(r,h) = πr2h.

For simplicity, let us consider a function with two variables and write z = f(x,y) to make explicit the value taken on by f at the general point (x,y). The variables x and y are independent with each other while z is dependent on both x and y. We will denote the derivative of z with respect to x as f_x and f_y for the derivative of z with respect to y. There are two rules in finding the partial derivatives of z = f(x,y):

1. To find f_x, take y as a constant and find the derivative of z with respect to x only.

2. To find f_y, take y as a constant and find the derivative of z with respect to y only.

There are many alternative notations for partial derivatives. For instance, instead of fx we can write f_1 or D_1f (to indicate differentiation with respect to the first variable) or \frac{\partial f}{\partial x}. But here \frac{\partial f}{\partial x} can’t be interpreted as a ratio of differentials.

## Notations for Partial Derivatives

There are many notations regarding partial derivatives. If z = f(x,y), then the most used notations for the partial derivatives of z are as follows;

### Partial derivative with respect to the variable x:

f_x(x,y)=f_x=\frac{\partial f}{\partial x} =\frac{\partial }{\partial x}f(x,y)=\frac{\partial z}{\partial x}=f_1=D_1f=D_xf

### Partial derivative with respect to the variable y:

f_y(x,y)=f_y=\frac{\partial f}{\partial y} =\frac{\partial }{\partial y}f(x,y)=\frac{\partial z}{\partial y}=f_2=D_2f=D_yf

### Example 1

Find the partial derivative of f(x,y) = x^2y^3 with respect to x..

Solution: Holding y constant and differentiating f with respect to x, we have

\frac{\partial }{\partial x}f(x,y)= \frac{\partial }{\partial x} (x^2 y^3 )= 2xy^3

### Example 2

Find the partial derivative of f(x,y) = x^2y^3 with respect to y.

For the partial derivative of f with respect to y, we take x as constant. Thus,

\frac{\partial }{\partial y}f(x,y)= \frac{\partial }{\partial y} (x^2 y^3 )= 3y^2 x^2

### Example 3

Solve the following partial derivative.

\frac{\partial }{\partial y}f(x,y)

where f(x,y) = \cos{x}+\cos{y}

For the partial derivative of f with respect to y, we take x as constant. Thus,

\frac{\partial }{\partial y}f(x,y)= \frac{\partial }{\partial y} ( \cos{x}+\cos{y})= 0-\sin{y}

Thus;

\frac{\partial }{\partial y} (\cos{x}+\cos{y})=-\sin{y}

### Example 4

Solve the following partial derivative.

\frac{\partial }{\partial x}f(x,y)

where f(x,y) = \cos{x}+\cos{y}

For the partial derivative of f with respect to x, we take y as constant. Thus,

\frac{\partial }{\partial x}f(x,y)= \frac{\partial }{\partial y} ( \cos{x}+\cos{y})= -\sin{x}+)

Thus;

\frac{\partial }{\partial y} (\cos{x}+\cos{y})=-\sin{x}

### Example 5

Differentiate:

\frac{\partial }{\partial z}(xy^2+\tan{xy^{\log{xy}}}\cdot e^{\cot{x^2y^2}})

Now, the function looks a little bit confusing and difficult. Since we are differentiating with respect to z, then the answer is zero since the function is independent of z.

Thus,