Partial Derivative (original) (raw)

Last Updated : 19 Jan, 2026

A Partial Derivative is when you take the derivative of a function with more than one variable, but focus on just one variable at a time, treating the others as constants.

For example, if f(x,y) = x2 + y3, the partial derivative with respect to x(\frac{∂f}{∂x})means:

• Treat y like a constant.
• Differentiate x2 + y3 as if it were just a function of x.

This gives:

\frac{∂f}{∂x} = 2x+0[As a derivative of a constant is 0]

Similarly, the partial derivative with respect to y (\frac{∂f}{∂y}) means:

• Treat x like a constant.
• Differentiate x2 + y3 as a function of y.

This gives:

\frac{∂f}{∂x} = 3y2.

We need partial derivatives because they help us understand how a function changes with respect to one variable at a time, especially when dealing with functions that depend on multiple variables.

**Note: Generally, **∂ this is the symbol of the partial derivative which is different from **d.

Partial Derivative Formula

Mathematically, consider a function f of dependent variables x, y, and z. Then the partial derivative of the function concerning x, y, and z can be written as

Partial derivative of a function with respect to x, keeping y and z constant:

**f x = \frac{∂f}{∂x}**= lim h→0 (f(x+h, y, z) - f(x, y, z))/h

Partial derivative of a function with respect to y, keeping x and z constant:

**f y **= \frac{∂f}{∂y} = lim h→0 (f(x, y+h, z) - f(x, y, z))/h

Partial derivative of a function with respect to z, keeping x and y constant:

**f z **=\frac{∂f}{∂z}**= lim h→0 (f(x, y, z+h) - f(x, y, z))/h

Partial Derivatives of Different Orders

Depending on the order of derivative required, the partial derivatives can vary. Let us see how:

First Order Partial Derivatives

Formula for calculating First Order Partial Derivatives is given by

fx = \frac {\partial f }{\partial x}and fy = \frac {\partial f }{\partial y}

For example considered above let's calculate the value.

f(x,y) = x2y+3y2

fx = 2xy + 0
fx (2,1) = 4

fy = x2+6y
fy(2,1) = 4+6 = 10

Second Order Partial Derivatives

Formula for calculating second Order Partial Derivatives is given by

fx' = \frac {\partial^2 f }{\partial x^2} and fy' = \frac {\partial^2 f }{\partial y^2}

For example considered above example

f'x = 𝛛(2xy)/𝛛x
f'x(2,1) = 2.y = 2

f'y = 𝛛(x2+6y)/𝛛y
f'y(2,1) = 6

Partial Differentiation

Partial differentiation refers to the process of calculating the partial derivative of a given function. Mathematically Partial Differentiation gives the slope of tangent drawn to the graph of the function at any point.

Consider a function f(x,y) = x2y + 3y2, we would like to see the first-order and second-order partial derivative of a function at x = 2 and y = 1.

**With respect to x: \frac{∂f}{∂x} = 2xy

• At x = 2 and y = 1: \frac{∂f}{∂x} = 2(2)(1) = 4

****With respect to y:**\frac{∂f}{∂y} = x2 + 6y

• At x = 2 and y = 1: \frac{∂f}{∂y} = (2)2 + 6(1) = 4 + 6 = 10

For second order partial derivatives:

**Second derivative with respect to x:: \frac{∂^2f}{∂x^2} = 2y

• x = 2 and y = 1: : \frac{∂^2f}{∂x^2}= 2(1) = 2

**Second derivative with respect to y: \frac{∂^2f}{∂y^2} = 6

• At x = 2 and y = 1: : \frac{∂^2f}{∂y^2} = 6

**Mixed partial derivative: \frac{∂^2f}{∂x∂y} = 2x

• At x = 2 and y = 1:\frac{∂^2f}{∂x∂y}= 2(2) = 4

Partial Derivative Rules

Let us see some rules used for calculating the partial derivative of a given function

Product Rule

This rule is used when a function is a product of two different functions i.e. u = f(x,y).g(x,y). According to the product rule, the partial derivative of this function will be,

• **u x **= (𝛛f(x,y)/𝛛x).g(x,y) + f(x,y).(𝛛g(x,y)/𝛛x)
• **u y **= (𝛛f(x,y)/𝛛y).g(x,y) + f(x,y).(𝛛g(x,y)/𝛛y)

Quotient Rule

This rule is used when a function is the quotient of two different functions i.e. u = f(x,y)/g(x,y). According to the quotient rule, the partial derivative of this function will be

• **u x **= (𝛛f(x,y)/𝛛x).g(x,y) - f(x,y).(𝛛g(x,y)/𝛛x)/(g(x,y)) 2
• **u y **= (𝛛f(x,y)/𝛛y).g(x,y) - f(x,y).(𝛛g(x,y)/𝛛y) /(g(x,y)) 2

Power Rule

This rule is used when a function is in the power of some number I.e u = (f(x,y))n. According to the power rule, the partial derivative of this function will be,

• u x **= n. |f(x,y)| n-1 (𝛛f(x,y)/𝛛x)
• **u y **= n. |f(x,y)| n-1 (𝛛f(x,y)/𝛛y)

Chain Rule

The chain rule is a tool used for calculating the derivative of a multivariable function. According to the chain rule of derivatives

**Chain Rule for One Independent Variable:

Let us consider two continuous functions u that are dependent on one variable t given by if x = g(t) andy=h(t). We have z = f(x, y) which is a differentiable function of x and y.

Then partial derivative

𝛛z/𝛛t = 𝛛z/𝛛x. 𝛛x/𝛛t + 𝛛z/𝛛y. 𝛛y/𝛛t

**Chain Rule for Two Independent Variables:

Let us consider two continuous functions u that are dependent on two variables u and v given by x = g (u, v) and y = h (u, v). We have z = f(x, y) which is a differentiable function of x and y.We can write f as z = f (g (u, v), h (u, v)). Then partial derivative

𝛛z/𝛛u = 𝛛z/𝛛x. 𝛛x/𝛛u + 𝛛z/𝛛y. 𝛛y/𝛛u and 𝛛z/𝛛v = 𝛛z/𝛛x. 𝛛x/𝛛v + 𝛛z/𝛛y. 𝛛y/𝛛v

Total Derivative vs Partial Derivative

Let us compare the total derivative and the partial derivative.

Applications of Partial Derivative

Various applications of partial derivative includes:

• Partial derivative is used in mathematical models that use complex equations. In such systems, the partial derivative helps to find the approximate solution of the system with efficiency.
• Partial derivatives are used in engineering-control systems. The control systems include dynamic systems that are complex to analyze with simple engineering tools. Therefore, partial derivates help to model and analyze such systems.
• Partial derivatives are used in chemical reactions to study the reaction kinetics of different reagents. They are mainly used to study reaction kinetics with changes in the concentration of different reagents and products.
• Partial derivatives are used in population dynamics to study how the population would change with changes in individual factors like birth rate, death rate, immigration, and calamities.
• Partial derivatives are used in the economics and finance sectors to predict future income. Modifications are done according to the results to maximize profits. For example, it helps to calculate a firm's output in terms of input factors like labor or capital.

Related Articles:

• Differential Equation
• Partial Differential Equation
• Solving Differential Equation
• Exact Differential Equation
• Read More about Derivative Rules

Partial Derivative Examples

**Example 1: Find the partial differential coefficient of the function xy 2 with respect to y where x 2 + xy + y 2 = 1.

**Solution:

Let z = xy2, we have to find the partial differential coefficient of z concerning y, that is, 𝛛z/𝛛y

We can write,

Let w = x2+ xy + y2 = 1

Differentiating both sides concerning y, we get

𝛛w/𝛛y = 0
⇒ 2xdx/dy + x + y.dx/dy+ 2y = 0
⇒ x + 2y = 0
⇒ x = -2y

f(x, y) = xy2
⇒ f(x,y) = (-2y).y2
⇒ f(x,y) = -2y3
⇒ 𝛛f(x,y)/𝛛y = (-6).y2

**Example 2: Find the partial differential coefficient of the function f(x,y,z) = x 2 y+ y 2 z+ xz with respect to x at x = 1, y = 2, z = 1.

**Solution:

f(x,y,z) = x2y+ y2z+ xz

Calculating partial derivative with respect to x, we will consider y and z to be constants ∴

𝛛f(x,y,z)/𝛛x = 2xy + 0 + z
⇒ 𝛛f(x,y,z)/𝛛x = 2xy +z

On putting the values x = 1, y = 2, z = 1
𝛛f(x,y,z)/𝛛x = 2.1.2 +1 = 5

**Example 3: Find the partial differential of the function f(x, y) = x 2 y + y 2 x with respect to x at x = 2 and y = 3.

**Solution:

**f(x, y) = x 2 y + y 2 x

Calculating partial derivative with respect to x, we will consider y and z to be constants ∴

𝛛f(x, y)/𝛛x = 2xy + y2

Putting the values x = 2, y = 3

{𝛛f(x, y)/𝛛x}(at x = 2 and y = 3) = 2.2.3 + (3)2 = 12 + 9 = 21

Practice Questions on Partial Derivative

Here are some problems for practice purposes.

Q1. Given, u = cos(x/y), x = et, y = t2, find δu/δx at t = 1. Verify your result by direct substitution.

Q2. Given, f(x, y) = exsin(y). Then evaluate δu/δy at x = 0 given x2+ y2 = 1.

Q3. Given, f(x, y) = ln(x/y).sin(y/x). Then evaluate δu/δy at x = 0 given x2+ y2 = 1.

Q4. Given, u = tan(x.y), x = sin(t), y = 3t2, find δu/δx at t = 1.Verify your result by direct substitution.