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Partial Derivative

In mathematics, sometimes the function depends on two or more than two variables. In this case, the derivative convert into the partial derivative since the function depends on several variables. Partial derivatives are used in vector calculus and differential geometry.

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Definition of Partial Derivative

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If we have a function f(x,y) i.e. a function which is depends on two variable x and y, where x and y are independent to each other. Then we say that the function f is partially depends on x and y. At this time if we calculate the derivative of f, then that derivative is called the partial derivative of f. If we differentiate f with respect to x, then taking y as a constant and if we differentiate f with respect to y, then taking x as a constant.

In mathematics, a partial derivative of a function of several variables is its derivative with respect to one of those variables, with the others held constant (as opposed to the total derivative, in which all variables are allowed to vary).

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 $\partial x$.

$\frac{\partial f}{\partial x}$ is also known as fx.

Partial Derivative Symbol

The partial derivative of a function f with respect to the variable x is variously denoted by $f'_x$, $f_x$, $\partial_x f$ or $\frac{\partial f}{\partial x}$
Here, $\partial$ is the symbol of partial derivative.

Partial Derivative of In

To find the partial derivative of function express in to form of "In", we follow the same procedure as finding the derivative of the function. But, this time when we calculate the partial derivative of the function with respect to one independent variable taking other as constant and follow the same thing with others.

Partial Derivative Examples

Given below are some of the examples on Partial Derivatives.

Solved Examples

Question 1: Determine the partial derivative of a function fx and fy: if f(x, y) is given by f(x, y) = tan(xy) + sin x
Solution:

Given function is f(x, y) = tan(xy) + sin x

Derivative of a function with respect to x is given as follows:

fx = $\frac{\partial f}{\partial x}$ = $\frac{\partial}{\partial x}$$[\tan (xy) + \sin x]$

= $\frac{\partial}{\partial x}$$ [\tan(xy)] + $$\frac{\partial}{\partial x}$ $[\sin x]$

= sec2 (xy). y + cos x

= y sec2(xy) + cos x

Now, Derivative of a function with respect to y. So, x is constant

fy = $\frac{\partial f}{\partial y}$ = $\frac{\partial}{\partial y}$$ [\tan (xy) + \sin x]$

= $\frac{\partial}{\partial y}$$ [\tan (xy)] + $$\frac{\partial}{\partial y}$$[\sin x]$

= sec2 (xy) x + 0

= x sec2 (xy)

Answer: fx = y sec2(xy) + cos x and fy = x sec2 (xy)



Question 2: If $f = x^{2}(y – z) + y^{2}(z – x) + z^{2}(x – y)$, prove that $\frac {\partial f} {\partial x}$ + $\frac {\partial f} {\partial y}$ + $\frac {\partial f} {\partial z}$$ + 0$
Solution:
Given, $f = x^{2} (y – z) + y^{2}(z – x) + z^{2}(x – y)$

To find $\frac {\partial f} {\partial x}$ ‘y and z’ are held constant and the resulting function of ‘x’ is differentiated with respect to ‘x’

$\frac {\partial f} {\partial x}$ = $2x(y - z) + 0(0 - 1) + 0(1 - 0)$

$\frac {\partial f} {\partial x}$ = $2xy - 2xz + 0 + 0$

$\frac {\partial f} {\partial x}$ = $2xy - 2xz$--------(i)

To find $\frac {\partial f} {\partial y}$ 'x and z’ is held constant and the resulting function of ‘y’ is differentiated with respect to ‘y’.

$\frac {\partial f} {\partial y}$ = $0(1 - z) + 2y(z - x) + 0(0 - 1)$

$\frac {\partial f} {\partial y}$ = $0 + 2yz - 2yx + 0$

$\frac {\partial f} {\partial y}$ = $2yz - 2yx$ -------(ii)

To find $\frac {\partial f} {\partial z}$ 'x and y’ is held constant and the resulting function of ‘z’ is differentiated with respect to ‘z’.

$\frac {\partial f} {\partial z}$ = $0(0 - 1) + 0(1 - 1) + 2z(x - y)$

$\frac {\partial f} {\partial z}$ = $0 + 0 + 2zx - 2zy$

$\frac {\partial f} {\partial z}$ = $2zx - 2zy$ ---------(iii)

Adding (i) , (ii) and (iii) we get

$\frac {\partial f} {\partial x}$ + $\frac {\partial f} {\partial y}$ + $\frac {\partial f} {\partial z}$ = $2xy - 2xz + 2yz - 2yx - 2zx - 2zy$

$\frac {\partial f} {\partial x}$ + $\frac {\partial f} {\partial y}$ + $\frac {\partial f} {\partial z}$ = $0$

Hence proved.



Partial Derivative Formula

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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

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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.

Second Partial Derivative

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As the partial derivatives of a function $f(x, y)$ are themselves functions of ‘x’ and ‘y’, they in turn may have partial derivatives.

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. We can define the second order partial derivatives as follows

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

The partial differentiation $f_{xy}$ and $f_{yx}$ are distinguished by the order on 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 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 (x 1 , y1) are the points of the function which satisfying equation (a), then
  1. If fxx < 0 and fyy < 0 then (x 1 , y1) is relative maximum point of the function.
  2. If fxx < 0 and fyy < 0 then (x 1 , y1) is relative minimum point of the function.

Solved Example

Question: 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

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

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

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

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

fxy = ab. eax cos (by)

(iv) fyx = (fy)x

fyx = (fy)x

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

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

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

fyx = ab. eax cos (by)

Partial Derivative Rules

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Just like ordinary derivatives, partial derivatives follows some rule like product rule, quotient rule, chain rule etc.

Partial Derivative Properties

If we have u = f(x,y) then, partial derivatives follows some rules as the ordinary derivatives.

Product Rule:

If u = f(x,y).g(x,y), then

          $u_{x}$ = $\frac{\partial u}{\partial x}$ = $g\left ( x,y \right )$$\frac{\partial f}{\partial x}$$ + f\left ( x,y \right )$$\frac{\partial g}{\partial x}$

And, $u_{y}$ = $\frac{\partial u}{\partial y}$ = $g\left ( x,y \right )$$\frac{\partial f}{\partial y}$$ + f\left ( x,y \right )$$\frac{\partial g}{\partial y}$

Quotient Rule:

If u = $\frac{f(x,y)}{g(x,y)}$, where g(x,y) $\neq$ 0 then,

      $u_{x}$ = $\frac{g\left ( x,y \right )\frac{\partial f}{\partial x}-f\left ( x,y \right )\frac{\partial g}{\partial x}}{\left [ g\left ( x,y \right ) \right ]^{2}}$

And, $u_{y}$ = $\frac{g\left ( x,y \right )\frac{\partial f}{\partial y}-f\left ( x,y \right )\frac{\partial g}{\partial y}}{\left [ g\left ( x,y \right ) \right ]^{2}}$

Power Rule:

If u = [f(x,y)]2 then, partial derivative of u with respect to x and y defined as

         $u_{x} = n\left [ f\left ( x,y \right ) \right ]^{n - 1} $$\frac{\partial f}{\partial x}$

And, $u_{y} = n\left [ f\left ( x,y \right ) \right ]^{n - 1} $$\frac{\partial f}{\partial y}$

Chain Rule:

If u = f(x,y) is a function where, x = (s,t) and y = (s,t) then by the chain rule, we can find the partial derivatives us and ut as:

$u_{s}$ = $\frac{\partial u}{\partial x}.\frac{\partial x}{\partial s} + \frac{\partial u}{\partial y}.\frac{\partial y}{\partial s}$
and $u_{t}$ = $\frac{\partial u}{\partial x}.\frac{\partial x}{\partial t} + \frac{\partial u}{\partial y}.\frac{\partial y}{\partial t}$   

Solved Examples

Question 1: f(x,y) = x2y + sin x + cos y
Solution:

Now, finding out fx first keeping y as constant

fx = $\frac{\partial f}{\partial x}$ = (2x) y + cos x + 0

= 2xy + cos x

when we keep y as constant cos y becomes a constant so its derivative becomes zero.

Similarly, finding fy

fy = $\frac{\partial f}{\partial y}$ = x2 + 0 + (-sin y)

= x2 - sin y



Question 2: xy + x2
Solution:
Lets find out the partial derivative with respect to x keeping y as constant.

$\frac{\partial f}{\partial x}$ = y + 2x

Here, while differentiating the xy term, we treat y as a constant. So, the differentiation of x is 1 which will simply get multiplied by y and the second term is x2 in terms of x only so its derivative is simply 2x.

Now, for the same problem we try to find out partial derivative with respect to y or fy

fy = $\frac{\partial f}{\partial y}$ = x + 0

treating x term as constant the second term x2 becomes a constant so its derivative with respect to y is 0.

fy = x



Mixed Partial Derivative

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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.

Solved Example

Question: 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’

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

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

To find $\frac {\partial f} {\partial y}$,

$\frac {\partial f} {\partial y}$ = $x^{3}(4y^{3}) - \sin (x)$

$\frac {\partial f} {\partial y}$ = $4x^{3}y^{3} - \sin (x)$

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

Partial Derivative Identities

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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 to total differentiation. Then the total partial derivative of u with respect to t is $\frac{df}{dt} = \frac{\partial f}{\partial x}\frac{dx}{dt} + \frac{\partial f}{\partial y}\frac{dy}{dt}$
  2. 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 $\frac{\partial f}{\partial v} = \frac{\partial f}{\partial x}\frac{\partial x}{\partial v}$
  3. 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 $\frac{\partial f}{\partial v} = \frac{\partial f}{\partial x}\frac{\partial x}{\partial v} + \frac{\partial f}{\partial y}\frac{\partial y}{\partial v}$

Solved Examples

Question 1: 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 2: If f(x,y) = 2x + 3y, where x = t and y = t2, find $\frac{df}{dt}$.
Solution:
We know that $\frac{df}{dt} = \frac{\partial f}{\partial x}\frac{dx}{dt} + \frac{\partial f}{\partial y}\frac{dy}{dt}$

Then, $\frac{\partial f}{\partial x}$ = 2, $\frac{\partial f}{\partial y}$ = 3, $\frac{dx}{dt}$ = 1, $\frac{dy}{dt}$ = 2t
So, we get
$\frac{df}{dt}$ = 2.1 + 3.(2t) = 2 + 3t

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