The partial derivative of functions is an important topic in calculus. If we have a function f(x,y), i.e., a function which depends on two variables, x and y, where x and y are independent of each other, then we say that the function f partially depends on x and y. The derivative of f is called the partial derivative of f. When we differentiate f with respect to x, then consider y as a constant, and when we differentiate f with respect to y, then consider x as a constant.

For example, suppose f is a function in x and y; then, it will be denoted by f(x,y). So, the partial derivative of f with respect to x will be ∂f/∂x, keeping y terms constant. Note that it is not dx, instead, it is ∂x.

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It can also be denoted as f’_x, f_x, ∂_xf or ∂f/∂x.

If f(x, y) is a function where f partially depends on x and y. Then if we differentiate f, with respect to x and y, then the derivatives are called the partial derivative of f with respect to x and y. The formula for the partial derivative of f with respect to x taking y as a constant,

\begin{array}{l} f_{x} = \frac{\partial f}{\partial x} = lim_{h \to 0} \frac{f (x + h, y) - f (x, y)}{h} \end{array}

And partial derivative of f with respect to y, taking x as a constant,

\begin{array}{l} f_{y} = \frac{\partial f}{\partial y} = lim_{h \to 0} \frac{f (x, y + h) - f (x, y)}{h} \end{array}

Partial Derivative Formulas and Identities

There are some identities for partial derivatives, as per the definition of the function.

1. If u = f(x, y) and both x and y are differentiable of t, i.e., x = g(t) and y = h(t), then the term differentiation becomes total differentiation.

2. The total partial derivative of u with respect to t is

\begin{array}{l} \frac{d f}{d t} = \frac{\partial f}{\partial x} \frac{d x}{d t} + \frac{\partial f}{\partial y} \frac{d y}{d t} \end{array}

3. If f is a function defined as f(x), where x(u, v) then

\begin{array}{l} \frac{\partial f}{\partial u} = \frac{\partial f}{\partial x} \frac{\partial x}{\partial u} \end{array}

and

\begin{array}{l} \frac{\partial f}{\partial v} = \frac{\partial f}{\partial x} \frac{\partial x}{\partial v} \end{array}

Suppose f = f(x,y) and y is an implicit function; it means y is itself a function of x, then

\begin{array}{l} \frac{d f}{d x} = \frac{\partial f}{\partial x} + \frac{\partial f}{\partial y} \frac{d y}{d x} \end{array}

4. If f(x,y) where x(u,v) and y(u,v), then

\begin{array}{l} \frac{\partial f}{\partial u} = \frac{\partial f}{\partial x} \frac{\partial x}{\partial u} + \frac{\partial f}{\partial y} \frac{\partial y}{\partial u} \end{array}

and

\begin{array}{l} \frac{\partial f}{\partial v} = \frac{\partial f}{\partial x} \frac{\partial x}{\partial v} + \frac{\partial f}{\partial y} \frac{\partial y}{\partial v} \end{array}

First Partial Derivative

If u = f(x,y), then the partial derivative of f with respect to x is defined as ∂f/∂x and denoted by

\begin{array}{l} \frac{\partial f}{\partial x} = lim_{δ x \to 0} \frac{f (x + δ x, y) - f (x, y)}{δ x} \end{array}

And partial derivative of f, with respect to y, is defined as ∂f/∂y and denoted by

\begin{array}{l} \frac{\partial f}{\partial y} = lim_{δ y \to 0} \frac{f (x, y + δ y) - f (x, y)}{δ y} \end{array}

These are called the first partial derivatives of f when we calculate the partial derivatives of f, with respect to x, treating y as a constant and vice versa.

Double Partial Derivative

Since the second-order partial derivative can be found by differentiating the first partial derivative, we can also call it the double partial derivative. The second-order partial derivatives can be defined as follows:

\begin{array}{l} \frac{\partial}{\partial x} (\frac{\partial f}{\partial x}) is denoted by \frac{\partial^{2} f}{\partial^{2} x} or f_{x x} or f_{x}^{2} \end{array}

and

\begin{array}{l} \frac{\partial}{\partial y} (\frac{\partial f}{\partial y}) is denoted by \frac{\partial^{2} f}{\partial y^{2} x} or f_{y y} or f_{y}^{2} \end{array}

and

\begin{array}{l} \frac{\partial}{\partial x} (\frac{\partial f}{\partial y}) is denoted by \frac{\partial^{2} f}{\partial x . \partial y} or f_{x y} \end{array}

and

\begin{array}{l} \frac{\partial}{\partial y} (\frac{\partial f}{\partial x}) is denoted by \frac{\partial^{2} f}{\partial y . \partial x} or f_{y x} \end{array}

The partial differentiation f_xy and f_yx are distinguished by the order in which ‘f’ is successively differentiated with respect to ‘x’ and ‘y’. In general, the two partial derivatives, f_xy and f_yx, need not be equal.

Second Partial Derivative Test

The necessary condition for the existence of a relative maximum and relative minimum of a function of two variables f(x,y) is

\begin{array}{l} \frac{\partial f}{\partial x} = 0 and \frac{\partial f}{\partial y} = 0 \dots . . (a) \end{array}

If (x₁, y₁) are the points of the function which satisfy equation (a), then

If f_xx < 0 and f_yy < 0, then (x₁, y₁) is the relative maximum point of the function.

If fxx < 0 and fyy < 0, then (x₁, y₁) is the relative minimum point of the function.

Mixed Partial Derivative

We can find out the mixed partial derivative or cross partial derivative of any function when the second-order partial derivative exists. If f(x,y) is a function with two independent variables, then we know that

\begin{array}{l} \frac{\partial f}{\partial x} = lim_{δ x \to 0} \end{array}

\begin{array}{l} \frac{f (x + δ x, y) - f (x, y)}{δ x} \end{array}

is the first partial derivative of f with respect to x.

Then,

\begin{array}{l} f_{x y} = \frac{\partial}{\partial y} (\frac{\partial f}{\partial x}) = \frac{\partial^{2} f}{\partial y \partial x} \end{array}

and

\begin{array}{l} f_{y x} = \frac{\partial}{\partial x} (\frac{\partial f}{\partial y}) = \frac{\partial^{2} f}{\partial x \partial y} \end{array}

The terms f_xy and f_yx are called the mixed or cross partial derivative of f.

Higher-Order Partial Derivatives

Second and higher-order partial derivatives are defined analogously to the higher-order derivatives of univariate functions. For the function f (x, y, . . .), the “own” second partial derivative with respect to x is simply the partial derivative of the partial derivative (both with respect to x).

The four second-order partial derivatives are as follows.

Let f (x, y) be a function of two variables, such that ∂f/∂x and ∂f/∂y both exist.

1) The partial derivative of ∂f/∂x w.r.t. ′x′ is denoted by ∂²f/∂x² or f_xx.

2) The partial derivative of ∂f/∂y w.r.t. ′y′ is denoted by ∂²f/∂y² or f_yy.

3) The partial derivative of ∂f/∂x w.r.t. ′y′ is denoted by ∂²f/(∂y ∂x) or f_xy.

4) The partial derivative of ∂f/∂y w.r.t. x is denoted by ∂²f/(∂y ∂x) or f_yx.

Also Read: Theorems on Differentiation

Solved Examples

Question 1: Consider the function f(x,y) = 5x⁴y²+ 6x²y³. Find f_x and f_y.

Solution:

Given f(x,y) = 5x⁴y²+ 6x²y³

Take y as constant and differentiate the given function, w.r.t x to find f_x.

f_x = 20x³y²+ 12xy³

Take x as constant and differentiate the given function, w.r.t y to find fy.

f_y = 5x⁴(2y) + 6x²(3y²) = 10x⁴y + 18x²y²

Question 2: Find the partial derivatives f_xy and f_yx of the function f(x,y) = x³y⁴ – y sin x

Solution:

To find f_xy and f_yx, first, we have to find f_x and f_y. So, ‘y’ is held constant, and the resulting function of ‘x’ is differentiated with respect to ‘x’

\begin{array}{l} \frac{\partial f}{\partial x} = y^{4} (3 x^{2}) - y . (\cos x) \end{array}

Or

\begin{array}{l} \frac{\partial f}{\partial x} = 3 x^{2} y^{4} - y . (\cos x) \end{array}

\begin{array}{l} \frac{\partial f}{\partial y} = x^{3} (4 y^{3}) - \sin (x) = 4 x^{3} y^{3} - \sin (x) \end{array}

Now,

\begin{array}{l} f_{x y} = \frac{\partial}{\partial y} [\frac{\partial f}{\partial y}] \end{array}

\begin{array}{l} = \frac{\partial}{\partial y} [3 x^{2} y^{4} - y c o s x] \end{array}

= 12x²y³ – cos x

Again,

\begin{array}{l} f_{y x} = \frac{\partial}{\partial x} [\frac{\partial f}{\partial x}] \end{array}

\begin{array}{l} = \frac{\partial}{\partial x} [4 x^{3} y^{3} - s i n x] \end{array}

= 12x²y³ – cos x

Question 3: If f(x,y) = xy and y = e^x, find df/dx.

Solution:

In this function, y is an implicit function, then we use

\begin{array}{l} \frac{d f}{d x} = \frac{\partial f}{\partial x} + \frac{\partial f}{\partial y} \frac{d y}{d x} \end{array}

Now,

\begin{array}{l} \frac{\partial f}{\partial x} = y, \end{array}

\begin{array}{l} \frac{\partial f}{\partial y} = x \end{array}

and

\begin{array}{l} \frac{d y}{d x} = e^{x} \end{array}

Then, df/dx = y + x e^x= e^x+ x e^x = e^x(1 + x)

Question 4: If f = e^ax sin (by), where ‘a’ and ‘b’ are real constants, find f_xx , f_yy , f_xy , f_yx

Solution:

Given, f = e^axsin (by),

(i) f_xx = (f_x)_x

To find f_x, ‘y’ is held constant, and the resulting function of ‘x’ is differentiated with respect to ‘x’.

f_xx = [a. e^ax sin (by)]

To find f_xx, once again, ‘y’ is held constant, and the resulting function of ‘x’ is differentiated with respect to ‘x’.

f_xx = a². e^ax sin (by)

(ii) f_yy = (f_y)_y

To find fy, ‘x’ is held constant, and the resulting function of ‘y’ is differentiated with respect to ‘y’.

f_yy = [b. e^ax cos (by)]

To find f_yy, once again, ‘x’ is held constant, and the resulting function of ‘y’ is differentiated with respect to ‘y’.

f_yy = – b². e^ax sin (by)

(iii) f_xy = (f_x)_y

To find f_x, ‘y’ is held constant, and the resulting function of ‘x’ is differentiated with respect to ‘x’.

f_xy = [a. e^ax sin (by)]

f_xy = ab. e^ax cos (by)

(iv) f_yx = (f_y)_x

f_yx = [b. e^ax cos (by)]

f_yx = ab. e^ax cos (by)

Question 5:

\begin{array}{l} If z = \tan^{- 1} (\frac{x}{y}), then z_{x} : z_{y} = \end{array}

Solution:

\begin{array}{l} \frac{\partial z}{\partial x} = \frac{1}{1 + \frac{x^{2}}{y^{2}}} . \frac{1}{y} = \frac{y}{x^{2} + y^{2}} \\ \frac{\partial z}{\partial y} = \frac{1}{1 + \frac{x^{2}}{y^{2}}} . (- \frac{x}{y^{2}}) \\ = - \frac{x}{x^{2} + y^{2}} \\ \frac{\partial z}{\partial x} : \frac{\partial z}{\partial y} = y : - x \\ i . e ., z_{x} : z_{y} = - y : x \end{array}

Question 6:

\begin{array}{l} If u = \frac{x + y}{x - y}, then \frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} = \end{array}

Solution:

\begin{array}{l} u = \frac{x + y}{x - y} \\ \frac{\partial u}{\partial x} = \frac{(x - y) . 1 - (x + y) . 1}{{(x - y)}^{2}} \\ = \frac{- 2 y}{{(x - y)}^{2}} \\ \frac{\partial u}{\partial y} = \frac{(x - y) .1 - (x + y) (- 1)}{{(x - y)}^{2}} \\ = \frac{2 x}{{(x - y)}^{2}} \\ \frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} = \frac{2 (x - y)}{{(x - y)}^{2}} \\ = \frac{2}{x - y} . \end{array}

Question 7:

\begin{array}{l} If u = \log (x^{2} + y^{2}), then \frac{\partial^{2} u}{\partial x^{2}} + \frac{\partial^{2} u}{\partial y^{2}} = \end{array}

Solution:

\begin{array}{l} u = \log (x^{2} + y^{2}) \\ \frac{\partial u}{\partial x} = \frac{1}{x^{2} + y^{2}} .2 x \\ \frac{\partial^{2} u}{\partial x^{2}} = \frac{(x^{2} + y^{2}) .2 - 2 x .2 x}{{(x^{2} + y^{2})}^{2}} \\ = \frac{2 (y^{2} - x^{2})}{{(x^{2} + y^{2})}^{2}} \\ \frac{\partial u}{\partial y} = \frac{1}{x^{2} + y^{2}} .2 y \\ \frac{\partial^{2} u}{\partial y^{2}} = \frac{(x^{2} + y^{2}) . 2 - 2 y .2 y}{{(x^{2} + y^{2})}^{2}} \\ = \frac{2 (x^{2} - y^{2})}{{(x^{2} + y^{2})}^{2}} \\ \frac{\partial^{2} u}{\partial x^{2}} + \frac{\partial^{2} u}{\partial y^{2}} = 0. \end{array}

Question 8:

\begin{array}{l} If u = \tan^{- 1} (\frac{x^{3} + y^{3}}{x - y}), then x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = \end{array}

Solution:
tanu is homogeneous in x, y of degree 2.

\begin{array}{l} x \frac{\partial}{\partial x} (\tan u) + y \frac{\partial}{\partial y} (\tan u) = 2 (\tan u) \\ x \sec^{2} u \frac{\partial u}{\partial x} + y \sec^{2} u \frac{\partial u}{\partial y} = 2 \tan u \\ x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = 2 \frac{\tan u}{\sec^{2} u} = 2 \sin u \cos u \\ = \sin 2 u . \end{array}

Question 9:

\begin{array}{l} If u = \log (x^{3} + y^{3} + z^{3} - 3 x y z), then (\frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} + \frac{\partial u}{\partial z}) | (x + y + z) | = \end{array}

Solution:

\begin{array}{l} u = \log (x^{3} + y^{3} + z^{3} - 3 x y z) \\ \frac{\partial u}{\partial x} = \frac{3 x^{2} - 3 y z}{x^{3} + y^{3} + z^{3} - 3 x y z} \\ \frac{\partial u}{\partial y} = \frac{3 y^{2} - 3 z x}{x^{3} + y^{3} + z^{3} - 3 x y z} \\ \frac{\partial u}{\partial z} = \frac{3 z^{2} - 3 x y}{x^{3} + y^{3} + z^{3} - 3 x y z} \\ \frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} + \frac{\partial u}{\partial z} = \frac{3 (x^{2} + y^{2} + z^{2} - x y - y z - z x)}{(x + y + z) (x^{2} + y^{2} + z^{2} - x y - y z - z x)} \\ = \frac{3}{x + y + z} \\ (x + y + z) (\frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} + \frac{\partial u}{\partial z}) = 3 \end{array}

Question 10:

\begin{array}{l} If u = x^{2} \tan^{- 1} \frac{y}{x} - y^{2} \tan^{- 1} \frac{x}{y}, then \frac{\partial^{2} u}{\partial x \partial y} = \end{array}

Solution:

\begin{array}{l} u = x^{2} \tan^{- 1} \frac{y}{x} - y^{2} (\frac{π}{2} - \tan^{- 1} \frac{y}{x}) \\ = (x^{2} + y^{2}) \tan^{- 1} \frac{y}{x} - \frac{π}{2} y^{2} \\ \frac{\partial u}{\partial y} = (x^{2} + y^{2}) \frac{1}{1 + \frac{y^{2}}{x^{2}}} . \frac{1}{x} + 2 y \tan^{- 1} \frac{y}{x} - π y \\ = x + 2 y \tan^{- 1} \frac{y}{x} - π y \\ \frac{\partial^{2} u}{\partial x \partial y} = 1 + 2 y \frac{1}{1 + \frac{y^{2}}{x^{2}}} . \frac{- y}{x^{2}} \\ = 1 - \frac{2 y^{2}}{x^{2} + y^{2}} = \frac{x^{2} - y^{2}}{x^{2} + y^{2}} \end{array}

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Frequently Asked Questions

Q1

When do we use partial derivatives?

We use partial derivatives when the function has more than one variable.

Q2

Give the chain rule for partial derivatives.

Suppose z = f(x, y), where x and y are functions of two variables, u and v. Then,
∂f/∂u = (∂f/∂x).(∂x/∂u) + (∂f/∂y).(∂y/∂u)
∂f/∂v = (∂f/∂x).(∂x/∂v) + (∂f/∂y).(∂y/∂v)

Q3

What is the difference between derivatives and partial derivatives?

We find the derivative of a function if the function has only one variable in it. Partial derivatives are used when a function z = f(x, y) has more than one variable.

Q4

How to denote the partial derivative of f, with respect to x?

The partial derivative of f, with respect to x, is denoted by ∂f/∂x.

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