# Partial Derivative of Functions

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 to 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, partial derivative of f with respect to x will be $\frac{\partial f}{\partial x}$ keeping y terms as constant. Note that, it is not dx, instead it is $\partial x$.

It can also be denoted as $f’_x$, $f_x$, $\partial_x f$ or $\frac{\partial f}{\partial x}$

## Partial Derivative 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 $\frac{df}{dt} = \frac{\partial f}{\partial x}\frac{dx}{dt} + \frac{\partial f}{\partial y}\frac{dy}{dt}$

3. If f is a function defined as f(x), where x(u,v) then $\frac{\partial f}{\partial u} = \frac{\partial f}{\partial x}\frac{\partial x}{\partial u}$ and $\frac{\partial f}{\partial v} = \frac{\partial f}{\partial x}\frac{\partial x}{\partial v}$

Suppose f = f(x,y) and y is a implicit function it means y is itself a function of x, then $\frac{df}{dx} = \frac{\partial f}{\partial x} + \frac{\partial f}{\partial y}\frac{dy}{dx}$

4. If f(x,y) where x(u,v) and y(u,v), then $\frac{\partial f}{\partial u} = \frac{\partial f}{\partial x}\frac{\partial x}{\partial u} + \frac{\partial f}{\partial y}\frac{\partial y}{\partial u}$ and $\frac{\partial f}{\partial v} = \frac{\partial f}{\partial x}\frac{\partial x}{\partial v} + \frac{\partial f}{\partial y}\frac{\partial y}{\partial v}$

## Partial Derivative Formula

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

$f_{x}$ = $\frac{\partial f}{\partial x}$ = $\lim_{h\rightarrow 0} \frac{f(x + h,y) – f(x,y)}{h}$

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

$f_{y}$ = $\frac{\partial f}{\partial y}$ = $\lim_{h\rightarrow 0} \frac{f(x,y + h) – f(x,y)}{h}$

First Partial Derivative

If u = f(x,y) is then the partial derivative of f with respect to x defined as $\frac{\partial f}{\partial x}$ and denoted by

$\frac{\partial f}{\partial x}$ = $\lim_{\delta x \rightarrow 0} \frac{f\left ( x + \delta x , y \right )-f\left ( x,y \right )}{\delta x}$

And, partial derivative of f with respect to y defined as $\frac{\partial f}{\partial y}$ and denoted by

$\frac{\partial f}{\partial y}$ = $\lim_{\delta y\rightarrow 0} \frac{f\left ( x,y+\delta y \right )-f\left ( x,y \right )}{\delta y}$

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 vise versa.

Double Partial Derivative

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

$\frac {\partial } {\partial x} (\frac {\partial f} {\partial x})$ is denoted by $\frac {\partial ^{2} f} {\partial ^{2} x}$ or $f_{xx}$ or $f_{x}^{2}$ $\frac {\partial } {\partial y} \left ( \frac {\partial f} {\partial y} \right )$ is denoted by $\frac {\partial ^{2} f} {\partial y^{2} x}$ or $f_{yy}$ or $f_{y}^{2}$ $\frac {\partial } {\partial x} \left ( \frac {\partial f} {\partial y} \right )$ is denoted by $\frac {\partial ^{2} f} {\partial x. \partial y}$ or $f_{xy}$ $\frac {\partial } {\partial y} \left ( \frac {\partial f} {\partial x} \right )$ is denoted by $\frac {\partial ^{2} f} {\partial y. \partial x}$ or $f_{yx}$

The partial differentiation fxy and fyx are distinguished by the order on 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 relative maximum and relative minimum of a function of two variables f(x,y) is

$\frac{\partial f}{\partial x}$ = 0 and $\frac{\partial f}{\partial y}$ = 0 ………………(a)

If (x1 , y1) are the points of the function which satisfying 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 of with two independent variables, then we know that

$\frac{\partial f}{\partial x}$ = $\lim_{\delta x \rightarrow 0}$$\frac{f\left ( x + \delta x , y \right ) – f\left ( x,y \right )}{\delta x}$ is the first partial derivative of f with respect to x.

Then, $f_{xy}$ = $\frac{\partial }{\partial y}\left ( \frac{\partial f}{\partial x} \right )$ = $\frac{\partial ^{2}f}{\partial y \partial x }$ and $f_{yx}$ = $\frac{\partial }{\partial x}\left ( \frac{\partial f}{\partial y} \right )$ = $\frac{\partial ^{2}f}{\partial x \partial y }$

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 $\frac{\partial f}{\partial x},\frac{\partial f}{\partial y}$ both exist.

1] The partial derivative of $\frac{\partial f}{\partial y}$ w.r.t. ′x′ is denoted by $\frac{{{\partial }^{2}}f}{\partial {{x}^{2}}}\text{ } \text\ or {{f}_{xx}}.$

2] The partial derivative of $\frac{\partial f}{\partial y}$ w.r.t. ′y′ is denoted by $\frac{{{\partial }^{2}}f}{\partial {{y}^{2}}} \text \ or {{f}_{yy}}.$

3] The partial derivative of  $\frac{\partial f}{\partial x}$ w.r.t. ′y′ is denoted by $\frac{{{\partial }^{2}}f}{\partial y\,\partial x} \text \ or \ {{f}_{xy}}.$

4] The partial derivative of $\frac{\partial f}{\partial y}$ w.r.t. x is denoted by $\frac{{{\partial }^{2}}f}{\partial y\partial x} \text \ or \ {{f}_{yx}}.$

Theorems On Differentiation

Methods Of Differentiation

### Solved Examples

Question 1: Consider the function f(x,y) = 5x4y2+6x2y. 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) = $x^{3}y^{4} – y\ \sin(x)$

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’

$\frac {\partial f} {\partial x}=y^{4}(3x^{2}) – y.\ (\cos x)$

Or $\frac {\partial f} {\partial x}=3x^{2} y^{4} – y. (\cos x)$ $\frac {\partial f} {\partial y}=x^{3}(4y^{3}) – \sin (x)=4x^{3}y^{3} – \sin (x)$

Now, fxy = $\frac{\partial}{\partial y} [\frac {\partial f} {\partial y}]$

= $\frac{\partial}{\partial y}$ [ 3x2y4 – y cos x]

= 12x2y3 – cos x

Again, fyx = $\frac{\partial}{\partial x} [\frac {\partial f} {\partial x}]$

= $\frac{\partial}{\partial x}$ [4x3y3 – sin x]

= 12x2y3 – cos x

Question 3: If f(x,y) = xy and y = ex, find $\frac{df}{dx}$.

Solution:

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

$\frac{df}{dx} = \frac{\partial f}{\partial x} + \frac{\partial f}{\partial y}\frac{dy}{dx}$

Now, $\frac{\partial f}{\partial x}$ = y, $\frac{\partial f}{\partial y}$ = x and $\frac{dy}{dx}$ = ex

Then, $\frac{df}{dx}$ = 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)]x

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)] y

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)] y

fxy = ab. eax cos (by)

(iv) fyx = (fy)x

fyx = [b. eax cos (by)] x

fyx = ab. eax cos (by)

Question 5: If $z={{\tan }^{-1}}\left( \frac{x}{y} \right)$, then ${{z}_{x}}:{{z}_{y}}=$

Solution:

$\frac{\partial z}{\partial x}=\frac{1}{1+\frac{{{x}^{2}}}{{{y}^{2}}}}.\frac{1}{y}=\frac{y}{{{x}^{2}}+{{y}^{2}}} \\\frac{\partial z}{\partial y}=\frac{1}{1+\frac{{{x}^{2}}}{{{y}^{2}}}}.\left( -\frac{x}{{{y}^{2}}} \right)\\=-\frac{x}{{{x}^{2}}+{{y}^{2}}} \\ \frac{\partial z}{\partial x}:\frac{\partial z}{\partial y}=y:-x \\i.e., {{z}_{x}}:{{z}_{y}}=-y:x$

Question 6: If $u=\frac{x+y}{x-y}$, then $\frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}=$

Solution:

$u=\frac{x+y}{x-y} \\\frac{\partial u}{\partial x}=\frac{(x-y)\,.\,1-(x+y)\,.\,1}{{{(x-y)}^{2}}}\\=\frac{-2y}{{{(x-y)}^{2}}} \\\frac{\partial u}{\partial y}=\frac{(x-y).1-(x+y)(-1)}{{{(x-y)}^{2}}} \\= \frac{2x}{{{(x-y)}^{2}}} \\\frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}=\frac{2(x-y)}{{{(x-y)}^{2}}}\\=\frac{2}{x-y}.$

Question 7: If $u=\log ({{x}^{2}}+{{y}^{2}}),$ then $\frac{{{\partial }^{2}}u}{\partial {{x}^{2}}}+\frac{{{\partial }^{2}}u}{\partial {{y}^{2}}}=$

Solution:

$u=\log ({{x}^{2}}+{{y}^{2}})\, \\\frac{\partial u}{\partial x}=\frac{1}{{{x}^{2}}+{{y}^{2}}}.2x \\\frac{{{\partial }^{2}}u}{\partial {{x}^{2}}}=\frac{({{x}^{2}}+{{y}^{2}}).2-2x.2x}{{{({{x}^{2}}+{{y}^{2}})}^{2}}}\\=\frac{2({{y}^{2}}-{{x}^{2}})}{{{({{x}^{2}}+{{y}^{2}})}^{2}}} \\\frac{\partial u}{\partial y}=\frac{1}{{{x}^{2}}+{{y}^{2}}}.2y \\\frac{{{\partial }^{2}}u}{\partial {{y}^{2}}}=\frac{({{x}^{2}}+{{y}^{2}})\,.\,2-2y.2y}{{{({{x}^{2}}+{{y}^{2}})}^{2}}}\\=\frac{2({{x}^{2}}-{{y}^{2}})}{{{({{x}^{2}}+{{y}^{2}})}^{2}}} \\\frac{{{\partial }^{2}}u}{\partial {{x}^{2}}}+\frac{{{\partial }^{2}}u}{\partial {{y}^{2}}}=0.$