CHAIN RULE AND IMPLICIT FUNCTION

Gandhinager Institute Of
Techonology
Active learning Project

The Chain Rule
In this section, we will learn about:
The Chain Rule and its application
in implicit differentiation.
PARTIAL DERIVATIVES

Recall that the Chain Rule for functions
of a single variable gives the following rule
for differentiating a composite function.
If y = f(x) and x = g(t), where f and g are differentiable functions,
then y is indirectly a differentiable function of t, and
dy dy dx
dt dx dt
=

For functions of more than one variable,
the Chain Rule has several versions.
› Each gives a rule for differentiating
a composite function.
The first version (Theorem 2) deals with
the case where z = f(x, y) and each of
the variables x and y is, in turn, a function
of a variable t.
This means that z is indirectly a function of t,
z = f(g(t), h(t)), and the Chain Rule gives a formula
for differentiating z as a function of t.

Suppose that z = f(x, y) is a differentiable function of x and y, where
x = g(t) and y = h(t) are both differentiable functions of t.
› Then, z is a differentiable function of t
and
Theorem 2
dz f dx f dy
dt x dt y dt
∂ ∂
= +
∂ ∂

A change of ∆t in t produces changes
of ∆x in x and ∆y in y.
› These, in turn, produce a change of ∆z in z.
Proof

Then, from Definition
we have:
where ε1 → 0 and ε2 → 0 as (∆x, ∆y) → (0,0).
› If the functions ε1 and ε2 are not defined at (0, 0),
we can define them to be 0 there.
Proof
1 2
f f
z x y x y
y y
ε ε
∂ ∂
∆ = ∆ + ∆ + ∆ + ∆
∂ ∂

Dividing both sides of this equation by ∆t,
we have:
∆z
∆t
=
∂f
∂x
∆x
∆t
+
∂f
∂y
∆y
∆t
+ε1
∆x
∆t
+ε2
∆y
∆t
Proof
If we now let ∆t → 0 , hen
∆x = g(t + ∆tt) – g(t) → 0
as g is differentiable and thus continuous.
Similarly, ∆y → 0.
This, in turn, means that ε1→ 0 and ε2→ 0.

Thus,
dz
dt
=lim
∆t→0
∆z
∆t
=
∂f
∂x
lim
∆t→0
∆x
∆t
+
∂f
∂y
lim
∆t→0
∆y
∆t
+ lim
∆t→0
ε1( )lim
∆t→0
∆x
∆t
+ lim
∆t→0
ε2( )lim
∆t→0
∆y
∆t
=
∂f
∂x
dx
dt
+
∂f
∂y
dy
dt
+0 ⋅
dx
dt
+0 ⋅
dy
dt
=
∂f
∂x
dx
dt
+
∂f
∂y
dy
dt
Proof

Since we often write ∂z/∂x in place of ∂f/∂x,
we can rewrite the Chain Rule in the form
dz z dx z dy
dt x dt y dt
∂ ∂
= +
∂ ∂

If z = x2
y + 3xy4
, where x = sin 2t and
y = cos t, find dz/dt when t = 0.
› The Chain Rule gives:
4 2 3
(2 3 )(2cos 2 ) ( 12 )( sin )
dz z dx z dy
dt x dt y dt
xy y t x xy t
∂ ∂
= +
∂ ∂
= + + + −
Example 1

•It’s not necessary to substitute the expressions for x and
y in terms of t.
• We simply observe that, when t = 0,
we have x = sin 0 = 0 and y = cos 0 = 1.
• Thus,
Example 1
0
(0 3)(2cos0) (0 0)( sin 0) 6
t
dz
dt =
= + + + − =

The derivative in Example 1 can be
interpreted as:
› The rate of change of z
with respect to t as
the point (x, y) moves
along the curve C
with parametric equations
x = sin 2t, y = cos t

In particular, when t = 0,
› The point (x, y) is (0, 1).
› dz/dt = 6 is the rate
of increase as we move
along the curve C
through (0, 1).

If, for instance, z = T(x, y) = x2
y + 3xy4
represents the temperature at the point (x,
y), then
› The composite function z = T(sin 2t, cos t)
represents the temperature at points on C
› The derivative dz/dt represents the rate at which
the temperature changes along C.

Suppose z = f(x, y) is a differentiable function of x and y, where x =
g(s, t) and y = h(s, t)
are differentiable functions of s and t.
› Then,
Theorem 3
∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂
= + = +
∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂
z z x z y z z x z y
s x s y s t x t y t

If z = ex
sin y, where x = st2
and y = s2
t, find ∂z/∂s and ∂z/∂t.
› Applying Case 2 of the Chain Rule,
we get the following results.
Example 1
2 2
2
2 2 2
( sin )( ) ( cos )(2 )
sin( ) 2 cos( )
x x
st st
z z x z y
s x s y s
e y t e y st
t e s t ste s t
∂ ∂ ∂ ∂ ∂
= +
∂ ∂ ∂ ∂ ∂
= +
= +

Example 1
2 2
2
2 2 2
( sin )(2 ) ( cos )( )
2 sin( ) cos( )
∂ ∂ ∂ ∂ ∂
= +
∂ ∂ ∂ ∂ ∂
= +
= +
x x
st st
z z x z y
t x t y t
e y st e y s
ste s t s e s t

Case 2 of the Chain Rule contains three types of variables:
› s and t are independent variables.
› x and y are called intermediate variables.
› z is the dependent variable.
Notice that Theorem 3 has one term
for each intermediate variable.
Each term resembles the one-dimensional
Chain Rule in Equation 1.

To remember the Chain Rule,
it’s helpful to draw a tree diagram,
as follows.
TREE DIAGRAM
We draw branches from the dependent variable z to the intermediate
variables
x and y to indicate that z is a function
of x and y.

Then, we draw branches from x and y
to the independent variables s and t.
› On each branch,
we write the
corresponding
partial derivative.

To find ∂z/∂s, we find the product of the partial derivatives along
each path from z to s and then add these products:
z z x z y
s x s y s
∂ ∂ ∂ ∂ ∂
= +
∂ ∂ ∂ ∂ ∂
Similarly, we find ∂z/∂t by using
the paths from z to t.

Now, we consider the general situation in
which a dependent variable u is a function
of n intermediate variables x1, . . . , xn.
Each of this is, in turn, a function of m
independent variables t1 , . . ., tm.
Notice that there are n terms—one for
each intermediate variable.
The proof is similar to that of Case 1.

Suppose u is a differentiable function of
the n variables x1, x2, …, xn and each xj
is a differentiable function of the m variables
t1, t2 . . . , tm.
Theorem 4
Then, u is a function of t1, t2, . . . , tm
and
for each i = 1, 2, . . . , m.
1 2
1 2
n
i i i n i
xx xu u u u
t x t x t x t
∂∂ ∂∂ ∂ ∂ ∂
= + +×××+
∂ ∂ ∂ ∂ ∂ ∂ ∂

Write out the Chain Rule for the case
where w = f(x, y, z, t)
and
x = x(u, v), y = y(u, v), z = z(u, v), t = t(u, v)
› We apply Theorem 4 with n = 4 and m = 2.
Example 1
•The figure shows the tree diagram.
• We haven’t written the derivatives on the branches.
• However, it’s understood that, if a branch leads from
y to u, the partial derivative for that branch is ∂y/∂u.

With the aid of the tree diagram, we can now write the required
expressions:
w w x w y w z w t
u x u y u z u t u
w w x w y w z w t
v x v y v z v t v
∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂
= + + +
∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂
∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂
= + + +
∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂
Example 1

The Chain Rule can be used to give
a more complete description of the process
of implicit differentiation that was introduced
in Sections 3.5 and 14.3

We suppose that an equation of
the form F(x, y) = 0 defines y implicitly
as a differentiable function of x.
› That is, y = f(x), where F(x, f(x)) = 0
for all x in the domain of f.
If F is differentiable, we can apply Case 1 of the
Chain Rule to differentiate both sides of the equation
F(x, y) = 0 with respect to x.
Since both x and y are functions of x,
we obtain:
0
F dx F dy
x dx y dx
∂ ∂
+ =
∂ ∂

However, dx/dx = 1.
So, if ∂F/∂y ≠ 0, we solve for dy/dx
and obtain:
x
y
F
Fdy x
Fdx F
y
∂
∂=− =−
∂
∂
Equation 6

To get the equation, we assumed F(x, y) = 0 defines y implicitly as a
function of x.
The Implicit Function Theorem, proved
in advanced calculus, gives conditions under which this assumption is
valid.

•The theorem states the following.
• Suppose F is defined on a disk containing (a, b), where F(a, b)
= 0, Fy(a, b) ≠ 0, and Fx and Fy
are continuous on the disk.
• Then, the equation F(x, y) = 0 defines y as
a function of x near the point (a, b) and
the derivative of this function is given by
Equation 6.

•Find y’ if x3
+ y3
= 6xy.
• The given equation can be written as:
F(x, y) = x3
+ y3
– 6xy = 0
• So, Equation 6 gives:
Example 1
2 2
2 2
3 6 2
3 6 2
x
y
Fdy x y x y
dx F y x y x
− −
=− =− =−
− −
Now, we suppose that z is given implicitly
as a function z = f(x, y) by an equation of
the form F(x, y, z) = 0.
This means that F(x, y, f(x, y)) = 0
for all (x, y) in the domain of f.

If F and f are differentiable, then we can use the Chain Rule to
differentiate the equation F(x, y, z) = 0 as follows:
0
F x F y F z
x x y x z x
∂ ∂ ∂ ∂ ∂ ∂
+ + =
∂ ∂ ∂ ∂ ∂ ∂
However,
So, that equation becomes
( ) 1 and ( ) 0x y
x x
∂ ∂
= =
∂ ∂
0
F F z
x z x
∂ ∂ ∂
+ =
∂ ∂ ∂

If ∂F/∂z ≠ 0, we solve for ∂z/∂x and obtain
the first formula in these equations.
› The formula for ∂z/∂y is obtained in a similar manner.
FF
z z yx
F Fx y
z z
∂∂
∂ ∂ ∂∂=− =−
∂ ∂∂ ∂
∂ ∂
Equations 7

Again, a version of the Implicit Function Theorem gives conditions
under which
our assumption is valid.

•This version states the following.
• Suppose F is defined within a sphere containing
(a, b, c), where F(a, b, c) = 0, Fz(a, b, c) ≠ 0,
and Fx, Fy, and Fz are continuous inside the sphere.
• Then, the equation F(x, y, z) = 0 defines z as
a function of x and y near the point (a, b, c),
and this function is differentiable, with partial derivatives given
by Equations 7.

Find ∂z/∂x and ∂z/∂y if
x3
+ y3
+ z3
+ 6xyz = 1
› Let F(x, y, z) = x3
+ y3
+ z3
+ 6xyz – 1
Example 1
Then, from Equations 7, we have:
2 2
2 2
2 2
2 2
3 6 2
3 6 2
3 6 2
3 6 2
x
z
y
z
Fz x yz x yz
x F z xy z xy
Fz y xz y xz
y F z xy z xy
∂ + +
=− =− =−
∂ + +
∂ + +
=− =− =−
∂ + +

CHAIN RULE AND IMPLICIT FUNCTION

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