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.
Also Read:
It can also be denoted as f’x, fx, ∂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,
\(\begin{array}{l}f_{x} = \frac{\partial f}{\partial x} = \lim_{h\rightarrow 0} \frac{f(x + h,y) – f(x,y)}{h}\end{array} \)
And partial derivative of f with respect to y, taking x as a constant,
\(\begin{array}{l}f_{y} = \frac{\partial f}{\partial y} = \lim_{h\rightarrow 0} \frac{f(x,y + h) – f(x,y)}{h}\end{array} \)
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
\(\begin{array}{l}\frac{df}{dt} = \frac{\partial f}{\partial x}\frac{dx}{dt} + \frac{\partial f}{\partial y}\frac{dy}{dt}\end{array} \)
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
\(\begin{array}{l}\frac{df}{dx} = \frac{\partial f}{\partial x} + \frac{\partial f}{\partial y}\frac{dy}{dx}\end{array} \)
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
\(\begin{array}{l}\frac{\partial f}{\partial x} = \lim_{\delta x \rightarrow 0} \frac{f\left ( x + \delta x , y \right )-f\left ( x,y \right )}{\delta x}\end{array} \)
And partial derivative of f, with respect to y, is defined as ∂f/∂y and denoted by
\(\begin{array}{l}\frac{\partial f}{\partial y} = \lim_{\delta y\rightarrow 0} \frac{f\left ( x,y+\delta y \right )-f\left ( x,y \right )}{\delta y}\end{array} \)
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:
\(\begin{array}{l}\frac {\partial } {\partial x} (\frac {\partial f} {\partial x})\ \text{is denoted by}\ \frac {\partial ^{2} f} {\partial ^{2} x}\ \text{or}\ f_{xx}\ \text{or}\ f_{x}^{2}\end{array} \)
and
\(\begin{array}{l}\frac {\partial } {\partial y} \left ( \frac {\partial f} {\partial y} \right )\ \text{is denoted by}\ \frac {\partial ^{2} f} {\partial y^{2} x}\ \text{or}\ f_{yy}\ \text{or}\ f_{y}^{2}\end{array} \)
and
\(\begin{array}{l}\frac {\partial } {\partial x} \left ( \frac {\partial f} {\partial y} \right )\ \text{is denoted by}\ \frac {\partial ^{2} f} {\partial x. \partial y}\ \text{or}\ f_{xy}\end{array} \)
and
\(\begin{array}{l}\frac {\partial } {\partial y} \left ( \frac {\partial f} {\partial x} \right )\ \text{is denoted by}\ \frac {\partial ^{2} f} {\partial y. \partial x}\ \text{or}\ f_{yx}\end{array} \)
The partial differentiation fxy and fyx are distinguished by the order in which ‘f’ is successively differentiated with respect to ‘x’ and ‘y’. In general, the two partial derivatives, fxy and fyx, 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
\(\begin{array}{l}\frac{\partial f}{\partial x} = 0\ \text{and}\ \frac{\partial f}{\partial y} = 0…..(a)\end{array} \)
If (x1, y1) are the points of the function which satisfy equation (a), then
If fxx < 0 and fyy < 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
\(\begin{array}{l}\frac{\partial f}{\partial x} = \lim_{\delta x \rightarrow 0}\end{array} \)
\(\begin{array}{l}\frac{f\left ( x + \delta x , y \right ) – f\left ( x,y \right )}{\delta x}\end{array} \)
is the first partial derivative of f with respect to x.
Then,
\(\begin{array}{l}f_{xy} = \frac{\partial }{\partial y}\left ( \frac{\partial f}{\partial x} \right )=\frac{\partial ^{2}f}{\partial y \partial x }\end{array} \)
and
\(\begin{array}{l}f_{yx} = \frac{\partial }{\partial x}\left ( \frac{\partial f}{\partial y} \right ) = \frac{\partial ^{2}f}{\partial x \partial y }\end{array} \)
The terms fxy and fyx 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 fxx.
2) The partial derivative of ∂f/∂y w.r.t. ′y′ is denoted by ∂2f/∂y2 or fyy.
3) The partial derivative of ∂f/∂x w.r.t. ′y′ is denoted by ∂2f/(∂y ∂x) or fxy.
4) The partial derivative of ∂f/∂y w.r.t. x is denoted by ∂2f/(∂y ∂x) or fyx.
Also Read: Theorems on Differentiation
Solved Examples
Question 1: Consider the function f(x,y) = 5x4y2 + 6x2y3. Find fx and fy.
Solution:
Given f(x,y) = 5x4y2 + 6x2y3
Take y as constant and differentiate the given function, w.r.t x to find fx.
fx = 20x3y2 + 12xy3
Take x as constant and differentiate the given function, w.r.t y to find fy.
fy = 5x4(2y) + 6x2(3y2) = 10x4y + 18x2y2
Question 2: Find the partial derivatives fxy and fyx of the function f(x,y) = x3y4 – y sin x
Solution:
To find fxy and fyx, first, we have to find fx and fy. So, ‘y’ is held constant, and the resulting function of ‘x’ is differentiated with respect to ‘x’
\(\begin{array}{l}\frac {\partial f} {\partial x}=y^{4}(3x^{2}) – y.\ (\cos x)\end{array} \)
Or
\(\begin{array}{l}\frac {\partial f} {\partial x}=3x^{2} y^{4} – y. (\cos x)\end{array} \)
\(\begin{array}{l}\frac {\partial f} {\partial y}=x^{3}(4y^{3}) – \sin (x)=4x^{3}y^{3} – \sin (x)\end{array} \)
Now,
\(\begin{array}{l}f_{xy} = \frac{\partial}{\partial y} [\frac {\partial f} {\partial y}]\end{array} \)
\(\begin{array}{l}=\frac{\partial}{\partial y}[ 3x^2y^4 – y\ cos\ x]\end{array} \)
= 12x2y3 – cos x
Again,
\(\begin{array}{l}f_{yx} = \frac{\partial}{\partial x} [\frac {\partial f} {\partial x}]\end{array} \)
\(\begin{array}{l}=\frac{\partial}{\partial x}[4x^3y^3 – sin\ x]\end{array} \)
= 12x2y3 – cos x
Question 3: If f(x,y) = xy and y = ex, find df/dx.
Solution:
In this function, y is an implicit function, then we use
\(\begin{array}{l}\frac{df}{dx} = \frac{\partial f}{\partial x} + \frac{\partial f}{\partial y}\frac{dy}{dx}\end{array} \)
Now,
\(\begin{array}{l}\frac{\partial f}{\partial x} = y,\end{array} \)
\(\begin{array}{l}\frac{\partial f}{\partial y} = x\end{array} \)
and
\(\begin{array}{l}\frac{dy}{dx} = e^x\end{array} \)
Then, df/dx = y + x ex = ex + x ex = ex (1 + x)
Question 4: If f = eax sin (by), where ‘a’ and ‘b’ are real constants, find fxx , fyy , fxy , fyx
Solution:
Given, f = eax sin (by),
(i) fxx = (fx)x
To find fx, ‘y’ is held constant, and the resulting function of ‘x’ is differentiated with respect to ‘x’.
fxx = [a. eax sin (by)]
To find fxx, once again, ‘y’ is held constant, and the resulting function of ‘x’ is differentiated with respect to ‘x’.
fxx = a2. eax sin (by)
(ii) fyy = (fy)y
To find fy, ‘x’ is held constant, and the resulting function of ‘y’ is differentiated with respect to ‘y’.
fyy = [b. eax cos (by)]
To find fyy, once again, ‘x’ is held constant, and the resulting function of ‘y’ is differentiated with respect to ‘y’.
fyy = – b2. eax sin (by)
(iii) fxy = (fx)y
To find fx, ‘y’ is held constant, and the resulting function of ‘x’ is differentiated with respect to ‘x’.
fxy = [a. eax sin (by)]
fxy = ab. eax cos (by)
(iv) fyx = (fy)x
fyx = [b. eax cos (by)]
fyx = ab. eax cos (by)
Question 5:
\(\begin{array}{l}\text{If}\ z={{\tan }^{-1}}\left( \frac{x}{y} \right),\ \text{then}\ {{z}_{x}}:{{z}_{y}}=\end{array} \)
Solution:
\(\begin{array}{l}\frac{\partial z}{\partial x}=\frac{1}{1+\frac{{{x}^{2}}}{{{y}^{2}}}}.\frac{1}{y}=\frac{y}{{{x}^{2}}+{{y}^{2}}} \\\frac{\partial z}{\partial y}=\frac{1}{1+\frac{{{x}^{2}}}{{{y}^{2}}}}.\left( -\frac{x}{{{y}^{2}}} \right)\\=-\frac{x}{{{x}^{2}}+{{y}^{2}}} \\ \frac{\partial z}{\partial x}:\frac{\partial z}{\partial y}=y:-x \\i.e., {{z}_{x}}:{{z}_{y}}=-y:x\end{array} \)
Question 6:
\(\begin{array}{l}\text{If}\ u=\frac{x+y}{x-y},\ \text{then}\ \frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}=\end{array} \)
Solution:
\(\begin{array}{l}u=\frac{x+y}{x-y} \\\frac{\partial u}{\partial x}=\frac{(x-y)\,.\,1-(x+y)\,.\,1}{{{(x-y)}^{2}}}\\=\frac{-2y}{{{(x-y)}^{2}}} \\\frac{\partial u}{\partial y}=\frac{(x-y).1-(x+y)(-1)}{{{(x-y)}^{2}}} \\= \frac{2x}{{{(x-y)}^{2}}} \\\frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}=\frac{2(x-y)}{{{(x-y)}^{2}}}\\=\frac{2}{x-y}.\end{array} \)
Question 7:
\(\begin{array}{l}\text{If}\ u=\log ({{x}^{2}}+{{y}^{2}}),\ \text{then}\ \frac{{{\partial }^{2}}u}{\partial {{x}^{2}}}+\frac{{{\partial }^{2}}u}{\partial {{y}^{2}}}=\end{array} \)
Solution:
\(\begin{array}{l}u=\log ({{x}^{2}}+{{y}^{2}})\, \\\frac{\partial u}{\partial x}=\frac{1}{{{x}^{2}}+{{y}^{2}}}.2x \\\frac{{{\partial }^{2}}u}{\partial {{x}^{2}}}=\frac{({{x}^{2}}+{{y}^{2}}).2-2x.2x}{{{({{x}^{2}}+{{y}^{2}})}^{2}}}\\=\frac{2({{y}^{2}}-{{x}^{2}})}{{{({{x}^{2}}+{{y}^{2}})}^{2}}} \\\frac{\partial u}{\partial y}=\frac{1}{{{x}^{2}}+{{y}^{2}}}.2y \\\frac{{{\partial }^{2}}u}{\partial {{y}^{2}}}=\frac{({{x}^{2}}+{{y}^{2}})\,.\,2-2y.2y}{{{({{x}^{2}}+{{y}^{2}})}^{2}}}\\=\frac{2({{x}^{2}}-{{y}^{2}})}{{{({{x}^{2}}+{{y}^{2}})}^{2}}} \\\frac{{{\partial }^{2}}u}{\partial {{x}^{2}}}+\frac{{{\partial }^{2}}u}{\partial {{y}^{2}}}=0.\end{array} \)
Question 8:
\(\begin{array}{l}\text{If}\ u={\tan }^{-1}\left(\frac{{{x}^{3}}+{{y}^{3}}}{x-y}\right),\ \text{then}\ x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}=\end{array} \)
Solution:
tanu is homogeneous in x, y of degree 2.
\(\begin{array}{l}x\frac{\partial }{\partial x}(\tan u)+y\frac{\partial }{\partial y}(\tan u)=2(\tan u) \\ x{{\sec }^{2}}u\frac{\partial u}{\partial x}+y{{\sec }^{2}}u\frac{\partial u}{\partial y}=2\tan u \\ x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}=2\frac{\tan u}{{{\sec }^{2}}u} = 2\sin u\cos u\\=\sin 2u.\end{array} \)
Question 9:
\(\begin{array}{l}\text{If}\ u=\log ({{x}^{3}}+{{y}^{3}}+{{z}^{3}}-3xyz), \text{then}\ \left( \frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}+\frac{\partial u}{\partial z} \right)|(x+y+z)| =\end{array} \)
Solution:
\(\begin{array}{l}u=\log ({{x}^{3}}+{{y}^{3}}+{{z}^{3}}-3xyz)\\ \frac{\partial u}{\partial x}=\frac{3{{x}^{2}}-3yz}{{{x}^{3}}+{{y}^{3}}+{{z}^{3}}-3xyz}\\ \frac{\partial u}{\partial y}=\frac{3{{y}^{2}}-3zx}{{{x}^{3}}+{{y}^{3}}+{{z}^{3}}-3xyz}\\ \frac{\partial u}{\partial z}=\frac{3{{z}^{2}}-3xy}{{{x}^{3}}+{{y}^{3}}+{{z}^{3}}-3xyz}\\ \frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}+\frac{\partial u}{\partial z}=\frac{3\,({{x}^{2}}+{{y}^{2}}+{{z}^{2}}-xy-yz-zx)}{(x+y+z)({{x}^{2}}+{{y}^{2}}+{{z}^{2}}-xy-yz-zx)}\\ =\frac{3}{x+y+z}\ \\ \,(x+y+z)\,\left(\frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}+\frac{\partial u}{\partial z} \right)=3\end{array} \)
Question 10:
\(\begin{array}{l}\text{If}\ u={{x}^{2}}{{\tan }^{-1}}\frac{y}{x}-{{y}^{2}}{{\tan }^{-1}}\frac{x}{y}, \text{then}\ \frac{{{\partial }^{2}}u}{\partial x\,\partial \,y}=\end{array} \)
Solution:
\(\begin{array}{l}u={{x}^{2}}{{\tan }^{-1}}\frac{y}{x}-{{y}^{2}}\left(\frac{\pi }{2}-{{\tan}^{-1}}\frac{y}{x} \right)\\=({{x}^{2}}+{{y}^{2}}){{\tan }^{-1}}\frac{y}{x}-\frac{\pi}{2}{{y}^{2}}\\ \frac{\partial u}{\partial y}=({{x}^{2}}+{{y}^{2}})\frac{1}{1+\frac{{{y}^{2}}}{{{x}^{2}}}}.\frac{1}{x}+2y{{\tan }^{-1}}\frac{y}{x}-\pi y\\=x+2y{{\tan}^{-1}}\frac{y}{x}-\pi y\\ \frac{{{\partial}^{2}}u}{\partial x\,\partial y}=1+2y\frac{1}{1+\frac{{{y}^{2}}}{{{x}^{2}}}}.\frac{-y}{{{x}^{2}}}\\=1-\frac{2{{y}^{2}}}{{{x}^{2}}+{{y}^{2}}}=\frac{{{x}^{2}}-{{y}^{2}}}{{{x}^{2}}+{{y}^{2}}}\end{array} \)
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Video Lessons
Application of Derivatives Important JEE Main Questions
Important Theorems of Differentiation for JEE
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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