Partial Derivative

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

f_x = \frac {\partial f }{\partial x}and f_y = \frac {\partial f }{\partial y}

For example considered above let's calculate the value.

f(x,y) = x²y+3y²

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

f_y = x²+6y
f_y(2,1) = 4+6 = 10

Second Order Partial Derivatives

Formula for calculating second Order Partial Derivatives is given by

f_x^' = \frac {\partial^2 f }{\partial x^2} and f_y^' = \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.

Solved Examples

Example 1: Find the partial differential coefficient of the function xy² with respect to y where x²+ xy + y²= 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) = x²y+ y²z+ 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) = x²y + y²x with respect to x at x = 2 and y = 3.

Solution:

f(x, y) = x²y + y²x

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

Here are some problems for practice purposes.

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

Q2. Given, f(x, y) = e^xsin(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.

Related Articles:

• Differential Equation 
• Partial Differential Equation
• Solving Differential Equation
• Exact Differential Equation
• Derivative Rules
• Application of Partial Derivatives