Partial Derivative

The function of an independent variable say y = f(x). Here, we learn about quantities which depend on more than one independent variable.

For example Area of a rectangle depends on its length(x) and width(y).
which can be expressed as A = f(x,y)

Volume of a parallel piped depends on its length l, width w and height h.

Hence Volume = f(l, w, h)

Hence “Partial Differentiation” deals with the differentiation of a function of many independent variables.

Difference Quotient

Let u be a function of two independent variables x and y.

i.e u = f( x, y )
Let us define the partial derivative of f with respect to x and y separately as follows, which is similar to the difference in quotient we write for functions of one independent variable.

  $\frac{\partial u}{\partial x}$

  $\frac{\partial u}{\partial y}$

[ $\partial$()  is the symbol of partial derivative usually read as “del” and it should not be written as  ‘$\delta$ ‘ (delta) ]

Partial Derivative Symbol

( First partial derivative )

The derivative of u with respect to x is called the partial derivative of u with respect to x,
and is denoted by ux or $\frac{\partial u}{\partial x}$ .


The derivative of u with respect to y is called the partial derivative of u with respect to y,
and is denoted by uy or $\frac{\partial u}{\partial y}$ .

Partial Derivative Examples

  Here are some partial derivative examples.

Example 1:
f(x,y) = sin(x2y2) find partial derivatives of f(x,y).

Solution :
Given f(x,y) = sin(x2y2)

Now, finding out fx first keeping y as constant
fx = $\frac{\partial f}{\partial x}$ = cos(x2 y2) $\times$ 2 x y2


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

Similarly, finding fy
fy = $\frac{\partial f}{\partial y}$ = cos(x2 y2) $\times$ 2 y x2



Example 2:
f(x , y) = xy + x2  find partial derivatives of f(x,y).

Solution:
Given f(x , y)=xy + x2

Let's, finding out fx first keeping y as constant
fx$\frac{\partial f}{\partial x}$ = y + 2x


Now, for the same problem we try to find out partial derivative with respect to y
fy = $\frac{\partial f}{\partial y}$ = x + 0

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

Partial Derivative Rules

Given a function of many independent variables, the derivative of this function with respect to a particular independent variable, keeping all the other independent variables as constants is the general principle of partial differentiation.

Second Partial Derivative

These are analogous with the higher order ordinary derivatives.

Let u = f( x, y).

We are aware of the first order partial derivatives , ux = $\frac{\partial u}{\partial x}$ and uy = $\frac{\partial u}{\partial y}$  
The second order partial derivatives will be,  uxx = $\frac{\partial}{\partial x}$ $(\frac{\partial u}{\partial x} )$$\frac{\partial^{2} u}{\partial x^{2}}$

         uyy = $\frac{\partial}{\partial y}$ $(\frac{\partial u}{\partial y} )$ =  $\frac{\partial^{2} u}{\partial y^{2}}$

Mixed Partial Derivative

It is the second order partial derivative which is written as follows:

           uyx = $\frac{\partial}{\partial x}$ $(\frac{\partial u}{\partial x} )$ =  $\frac{\partial^{2} u}{\partial y \partial x}$
           uxy  = $\frac{\partial}{\partial x}$ $(\frac{\partial u}{\partial x} )$ =  $\frac{\partial^{2} u}{\partial x \partial y}$

It is very important to note that, $\frac{\partial^{2} u}{\partial y \partial x}$$\frac{\partial^{2} u}{\partial x \partial y}$
uxy = uyx [ This is also called as Cross Partial Derivatives ]

Partial Derivative Chain Rule

If  u = f(x,y) and if each of x and y are represented parametrically, then we use the chain rule as follows.
 
if x = x(t) and y = y(t)

then $\frac{dx}{dt}$ = x ' (t) and  $\frac{dy}{dt}$ = y ' (t)

The partial derivatives will be, ux = $\frac{du}{dx}$ . $\frac{dx}{dt}$ and uy$\frac{du}{dy}$  . $\frac{dy}{dt}$ 

Partial Derivative Product Rule

When u = f( x, y) . g(x,y) where the given function u is expressed as the product of two functions f and g, we use the same rule as that of the definite derivative, at the same time we apply the rule of partial derivative also, during which we retain the other independent variable as a constant.
 
Example : If u = ( 3x + 5y) ( 2x2 – 4 y2 )

then,

$\frac{du}{dx}$  = ( 3x + 5y ) $\frac{d}{dx}$  (2x2 – 4 y2 ) + ( 2x2 – 4 y2$\frac{d}{dx}$ ( 3x + 5y ) [ we treat y as a constant ]
   
= ( 3x + 5y ) ( 2. 2x) + ( 2x2 – 4 y2 ) ( 3)
  
= 4x ( 3x + 5y ) + 3 (2x2 – 4 y2 )

Find The Indicated Partial Derivative

You could see how to find the indicated partial derivative:  

Solved Examples

Question 1: Use the Chain Rule to find the indicated partial derivatives.

R = ln $(u^{2} + v^{2} + w^{2})$
u = x + 2y, 
v = 7x - y, 
w = 3xy; when x = y = 4.
Solution:
 
Given R = ln $ (u^{2} + v^{2} + w^{2}) $
u = x + 2y
v = 7x - y
w = 3xy

When x = y = 4:
u = 4 + 2 $\times$ 4 = 12
v = 7 $\times$ 4 - 4 = 24
w = 3 $\times$ 4 $\times$ 4 = 48

$\frac{\partial R}{\partial x}$ = ($\frac{\partial R}{\partial u}$ $\times$ $\frac{\partial u}{\partial x}$) + ($\frac{\partial R}{\partial v}$ $\times$ $\frac{\partial v}{\partial x}$) + ($\frac{\partial R}{\partial w}$ $\times$ $\frac{\partial w}{\partial x}$)

$\frac{\partial R}{\partial x}$ = $\frac{2u}{(u^{2}+v^{2}+w^{2})(1)}$ + $\frac{2v}{(u^{2}+v^{2}+w^{2})(7)}$ + $\frac{2w}{(u^{2}+v^{2}+w^{2})(3y)}$

$\frac{\partial R}{\partial x}$ = $\frac{2 \ \times \ 12}{(12^{2} \ + \ 24^{2} \ + 48^{2})(1)}$ + $\frac{2 \ \times \ 24}{(12^{2} \ + \ 24^{2} \ + 48^{2})(7)}$ + $\frac{2 \ \times \ 48}{(12^{2} \ + \ 24^{2} \ + 48^{2})(3 \ \times \ 4)}$


$\frac{\partial R}{\partial x}$ = $\frac{24}{3024 \times 1}$ + $\frac{48}{3024 \times 7}$ + $\frac{96}{3024 \times 12}$

$\frac{\partial R}{\partial x}$ = $\frac{24}{3024}$ + $\frac{336}{3024}$ + $\frac{1152}{3024}$

$\frac{\partial R}{\partial x}$ = $\frac{1512}{3024}$

$\frac{\partial R}{\partial x}$ = $\frac{1}{2}$



$\frac{\partial R}{\partial y}$ = $(\frac{\partial R}{\partial u} \times \frac{\partial u}{\partial y}) + (\frac{\partial R}{\partial v} \times \frac{\partial v}{\partial y}) + (\frac{\partial R}{\partial w} \times \frac{\partial w}{\partial y})$


$\frac{\partial R}{\partial y}$ = $\frac{2u}{(u^{2}+v^{2}+w^{2})(2)}$ + $\frac{2v}{(u^{2}+v^{2}+w^{2})(-1)}$ + $\frac{2w}{(u^{2}+v^{2}+w^{2})(3x)}$


$\frac{\partial R}{\partial y}$ = $\frac{2 \ \times \ 12}{(12^{2} \ + \ 24^{2} \ + 48^{2})(2)}$$\frac{2 \ \times \ 24}{(12^{2} \ + \ 24^{2} \ + 48^{2})(-1)}$ + $\frac{2 \ \times \ 48}{(12^{2} \ + \ 24^{2} \ + 48^{2})(3 \ \times \ 4)}$

$\frac{\partial R}{\partial y}$ = $\frac{24}{3024 \times 2}$ + $\frac{48}{3024 \times (-1)}$ + $\frac{96}{3024 \times 12}$

$\frac{\partial R}{\partial y}$ = $\frac{48}{3024}$ - $\frac{48}{3024}$ + $\frac{1152}{3024}$

$\frac{\partial R}{\partial y}$ = $\frac{1152}{3024}$

$\frac{\partial R}{\partial y}$ = $\frac{8}{21}$
 

Question 2: Use the Chain Rule to find the indicated partial derivatives.
Y = w tan-1(uv),
u = r + s,
v = s + t,
w = t + r;

$\frac{\partial Y}{\partial r}$,

$\frac{\partial Y}{\partial s}$,

$\frac{\partial Y}{\partial t}$


when r = 0, s = 1, t = 2

Solution:
 
Given Y = w tan-1(uv),
u = r + s,
v = s + t,
w = t + r;

$\frac{\partial Y}{\partial r}$ = $(\frac{\partial Y}{\partial u})(\frac{\partial u}{\partial r}) + (\frac{\partial Y}{\partial v})(\frac{\partial v}{\partial r}) + (\frac{\partial Y}{\partial w})(\frac{\partial w}{\partial r})$

$\frac{\partial Y}{\partial r}$ = $(\frac{w \times v}{(1 + (uv)^{2})})$(1) + $(\frac{w \times u}{(1 + (uv)^{2})})$(0) + (tan-1(uv))(1)

$\frac{\partial Y}{\partial r}$
= $\frac{w \times v}{(1 + (uv)^{2})} + tan-1(uv)


$\frac{\partial Y}{\partial s}$ = $(\frac{\partial Y}{\partial u})(\frac{\partial u}{\partial s}) + (\frac{\partial Y}{\partial v})(\frac{\partial v}{\partial s}) + (\frac{\partial Y}{\partial w})(\frac{\partial w}{\partial s}) $


$\frac{\partial Y}{\partial s}$ = $(\frac{w \times v}{(1 + (uv)^{2})})$ (1) + $(\frac{w \times u}{(1 + (uv)^{2})})$ (1) + (tan-1(uv))(0)


$\frac{\partial Y}{\partial s}$
= $\frac{w \times v}{(1 + (uv)^{2})} + \frac{w \times u}{(1 + (uv)^{2})})$


$\frac{\partial Y}{\partial s}$
= $\frac{w(v + u)}{1 + (uv)^{2}} $


$\frac{\partial Y}{\partial t}$ = $(\frac{\partial Y}{\partial u})(\frac{\partial y}{\partial t})$ + $(\frac{\partial Y}{\partial v})(\frac{\partial v}{\partial t})$ + $(\frac{\partial Y}{\partial w})(\frac{\partial w}{\partial t})$

$\frac{\partial Y}{\partial t}$ = $(\frac{w \times v}{(1 + (uv)^{2}})$(0) + $(\frac{w \times u}{(1 + (uv)^{2}})$ (1) + (tan-1(uv))(1)

$\frac{\partial Y}{\partial t}$ = $\frac{w \times u}{1 + uv^{2}}$ + tan-1(uv)

Evaluate u, v, and w at r = 0, s = 1, t = 2:
u = 0 + 1 = 1
v = 1 + 2 = 3
w = 2 + 0 = 2

Evaluate the partials at r = 0, s = 1, t = 2:

$\frac{\partial Y}{\partial r}$ = $\frac{2 \times 3)}{(1 + (1 \times 3)^{2})}$ + tan-1(1 $\times$ 3)

$\frac{\partial Y}{\partial r}$ = $\frac{3}{5}$ + tan-1(3)


$\frac{\partial Y}{\partial s}$ = $\frac{2(3 + 1)}{(1 + (1 \times 3)^{2})}$

$\frac{\partial Y}{\partial s}$ = $\frac{4}{5}$


$\frac{\partial Y}{\partial t}$ = $\frac{2 \times 1)}{(1 + (1 \times 3)^{2})}$ + tan-1(1 $\times$ 3)

$\frac{\partial Y}{\partial t}$ = $\frac{1}{5}$ + tan-1 (3)

 

Inflection Point

For any function u = f(x,y) the critical points are obtained when ux = 0 and uy = 0.
 
If uxx = uyy  = 0, at the points where ux = uy = 0 then those points are called the points of inflection.

Linear Approximation

 Let u = f(x,y) be the given surface. 

The equation of the tangent plane to the graph of the function f of two variables at the point a, b, f(a,b)) is,
z = f(a, b) + fx (a,b) (x – a) + fy ( a, b) ( y – b), which is a linear function.

Hence the function of the form,

f( x, y) = f(a, b) + fx (a,b) (x – a) + fy ( a, b) ( y – b) is called the linear approximation or the tangent plane approximation of f at (a, b).

Similarly, for  a function of 3 variables, x , y, z,

f( x, y, z) = f(a, b) + fx (a,b) (x – a) + fy ( a, b) ( y – b) + fz ( a, b) ( z – b) is the Linear approximation.

Newton Raphson Method

This locates  an approximate real root of the given equation and improves its accuracy by an iterative process.

Let f(x) = 0, be the given equation and x0 be an approximate root of the equation. If h is a small correction applied to the root, then x0 + h is the exact root and we try to find h such that f (x0 + h ) = 0.

The first approximation to the root x0 is given by, x1 = x0 + h

(i.e) x1 = x0$\frac{f(x_{0})}{f'(x_{0})}$  , provided ${f}'(x_{0})$ $\neq$ 0

The second approximation is obtained by replacing x0  by x1.

(i.e) x2 = x0 –  $\frac{f(x_{1})}{f'(x_{1})}$

In general, xn+1 = xn –  $\frac{f(x_{n})}{f'(x_{n})}$

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