This document discusses the total derivative and methods for finding derivatives of functions with multiple variables. The total derivative expresses the total differential of a function u with respect to time t as the sum of the partial derivatives of u with respect to each variable x1, x2,...xn, multiplied by the rate of change of that variable with respect to time. The chain rule is used to take derivatives of composite functions, where the output of one function is an input to another. The derivative is expressed as the product of the partial derivatives of each nested function. Derivatives can also be taken for implicit functions, where not all variables can be solved for explicitly. The derivative of one variable with respect to another in an

1. Total Derivative
(A) u f(x 1 , x 2 , x 3 ...., x n ) and u has continuous partial derivatives f x & f y
.
= = = = =
Here x i x i (t) where i 0,1,2,3,4.....n [ x 1 x 1 (t), x 2 x 2 (t)... x n x n
(t) ]
So indirectly u f[x (t), x (t), x (t),..., x (t)]
And the total differential coefficient OR Total derivative is given by :
dx
i
i =
n
= ¶
å
i
u
du
=
1
i n
du = ¶
u
å
i i
i
i
dx
x
dt
x
dt
du
dt
*
=
=
¶
*
¶
=
=
1
1 2 3 n
........ (1)
(B) Equation (1) can be written in differential form by eliminating 'dt' as

2. Total Derivative
x
1
= ¶
(C) By putting t x and making 1 ,
we get a total differential coefficient of u with respect to x :
dx
u
u
du
*
+ ¶
¶
= ¶
1 1 2 1
x x du
= * = =
If u e , where sin(t) & y t , find
y y
2 2
x e t x e t
dy
u
dx
u
Ans du
* = * * + * *
¶
= ¶
. 2 cos( ) (3 )
t e t t e t
= * * * + * * *
2 sin( ) cos( ) 3 sin ( )
sin( ) ( )(2 cos( ) 3 sin( ))
Example :
3 2
2 3 2
2 y 3
1
1
1
3
t e t t t
dt
y
dt
x
dt
dt
dx
x
x
dx
x
t
t t
i n i
i i
= * * + *
* + ¶
¶
¶
=
¶
å=
=

3. Total Derivative
Example : Find the total derivative of f(x, y) 3 with respect to ,
x x x
2 1/ 2
2 3sin 1
3
given that sin ( )
x y f
f
= + ¶
¶
Ans. 2 3 , 3
1
dy
dx
f
=
= ¶
-
+ ¶
¶
*
¶
y
dy
dx x
f
x
2 3 3 1
2 1/ 2
1
2 1/ 2
(1 )
(1 )
(1 )
x
x
x y x
df
dx
x
y
x
y x
x x y x
-
= + +
-
= + +
=
¶
¶
=
= + * *
-
-

4. The Chain Rule
If p f(x, y,z) & x (u, v,w), y y(u,v,w) & z z(u, v,w), then the composite function
z
z
* ¶
¶
* ¶
¶
* ¶
¶
w
p
p
p
z
y
* ¶
¶
y
y
* ¶
¶
* ¶
¶
w
p
p
p
y
x
* ¶
¶
x
x
* ¶
¶
* ¶
¶
w
p
p
p
x
p
¶
p
p
¶
¶
w
z
v
z
v
y
v
x
v
u
z
u
y
u
x
u
x
¶
+ ¶
¶
+ ¶
¶
= ¶
¶
¶
+ ¶
¶
+ ¶
¶
= ¶
¶
¶
+ ¶
¶
+ ¶
¶
= ¶
¶
=
= = = =
p f[x(u, v,w), y(u,v,w), z(u, v,w)] has partial derivatives with respect to u, v,w as follows :

5. The Chain Rule
= = + = +
u f(r,s), r x at, s y bt & x, y & t are the independent variables, then show that :
r
+ ¶
¶
Ans. 1 0 ...................(1)
r
+ ¶
¶
r
u
+ ¶
¶
= ¶
¶
u
= ¶
¶
u
= ¶
¶
= ¶
¶
* ¶
¶
* ¶
¶
a u
x
u
u
u
b
y
u
b u
u
u
u
¶
u
t
u
* = ¶
¶
u
u
u
* = ¶
¶
0 1 ....................(2)
a u
r
b
s
u
s
s
s
t
s
t
r
t
u
u
s
s
r
y
s
y
r
y
r
s
r
x
s
x
r
x
y
a u
x
u
t
* + ¶
¶
*
¶
¶
= ¶
¶
* + ¶
¶
*
¶
= ¶
¶
+ ¶
¶
= ¶
¶
¶
* + ¶
¶
* ¶
¶
* ¶
¶
¶
¶
* + ¶
¶
* ¶
¶
* ¶
¶
¶
¶
¶
= ¶
¶
From (1), (2) &(3)
.............................(3)
Example :

6. Derivative Of Implicit Functions
Sometimes in a function that depends on more than one variable,
A specific variable cannot be expressed as a functions of the other variable
2 2 2 2
x y y x x y e abc
+ + + = =
example : constant
This type of functions is called Implicit Function & they have a specific formula for differentition
x
f
- for any two varibles on which the function depends
¶
In the above example.... y
x
where & are the partial differentiation of function 'f ' with respect to x & y respectively.
¶
Here there is also another method of carrying out y
. First differentiate both sides the whole equation
x
with respect to 'x'( i.e. whenever we differentiate a term having both 'x' & ' y'
¶
& whenever we encounter ' a differentiation of y with respect to x ' , leave it as y
¶
¶
as common and shift whole terms we get as product with y
x
other side as denominator .
to the
x
Now collect all terms of y
)
x
first differntiate considering 'x' as constant & then by considering' y' as constant
¶
¶
¶
¶
=
¶
x y
y
xy
f f
f

7. Derivative Of Implicit Functions
.
ö
x y a
x y a
f
f y x y y
f x x x y
y x y y
-
+ * ¶
¶
y
y x
y x y y x x x y
y x y y
= - * + *
log( )
log( )
Example :
¶
Find y
y x b
y x b
Ans. Here constant
1
¶
¶
y -
x
x
y
y x
= * + *
¶
y -
1
x
-
1
1
æ
= - * + *
y x
log( ) log( ) y
y
¶
So.... y
¶
which again gives on simplifying y
x
0(differentiation of a constant)
x
x
By another method it becomes :
log( )
log( )
x
log( )
y
log( )
x
-
x
when
x
1
1
1 1
ö
÷ ÷ø
æ
ç çè
* + *
¶
=
¶
* + * + * ¶
÷ ÷ø
ç çè
* + *
¶
¶
= * + *
¶
=
¶
+ = =
+ =
¶
-
-
- -
y x
y x y x
x x x y
x x x y
f