Section 11.
5
The Chain Rule
(1) The Chain Rule for 2-Variable Functions,
(2) The Chain Rule for Functions,
(3) Implicit Differentiation with the Chain Rule.
MATH 127 (Section 11.5) The Chain Rule The University of Kansas 1 / 11
The Chain Rule
If y = f (x) and x = g (t), then the composition y = (f ◦ g )(t) = f (g (t))
is a function of t. In this case, we have the chain rule for differentiation
dy dy dx
y 0 (t) = f 0 (g (t))g 0 (t) or =
dt dx dt
We will now discuss the chain rule for functions of more variables.
Suppose that z = f (x, y ), x = x(t), and y = y (t) are differentiable
functions. Then z = f (x(t), y (t)) is a differentiable function of t and
dz ∂z dx ∂z dy
= +
dt ∂x dt ∂y dt
dz
= fx (x(t), y (t))x 0 (t) + fy (x(t), y (t))y 0 (t)
dt
MATH 127 (Section 11.5) The Chain Rule The University of Kansas 2 / 11
The Chain Rule (Case I)
Suppose that z = f (x, y ), x = x(t), and y = y (t) are differentiable
functions. Then z = f (x(t), y (t)) is a differentiable function of t and
dz ∂z dx ∂z dy
= + .
dt ∂x dt ∂y dt
The derivative of z with respect to t can be calculated as
dz f (x(t + ∆t), y (t + ∆t)) − f (x(t), y (t))
= lim
dt ∆t→0 ∆t
Note that ∆x → 0 and ∆y → 0 as ∆t → 0 because x and y are
differentiable, hence continuous, functions of t.
f (x + ∆x, y + ∆y ) − f (x, y + ∆y ) ∆x f (x, y + ∆y ) − f (x, y ) ∆y
= +
∆x ∆t ∆y ∆t
Letting ∆t → 0, we have
dz dz ∂z dx ∂z dy
= fx (x, y ) x 0 (t) + fy (x, y ) y 0 (t) = = +
dt dt ∂x dt ∂y dt
MATH 127 (Section 11.5) The Chain Rule The University of Kansas 3 / 11
Example
dz
If z = yx e xy , x = ln(t), and y = sin(t), find when t = π2 .
dt
Solution: The Chain Rule states that
dz ∂z dx ∂z dy
= +
dt ∂x
dt ∂y dt
1 xy xy 1 −x xy x 2 xy
= ye + xe t + y2 e + y e cos(t)
π π π
For t = 2 we have x = ln 2 and y = sin 2 = 1. Thus,
dz π
= 1 + ln
dt t= π 2
2
MATH 127 (Section 11.5) The Chain Rule The University of Kansas 4 / 11
Suppose z = f (x, y ) and let x = g (s, t) and y = h(s, t) be differentiable
2-variable functions in s and t. Then z = f (x, y ) can be viewed as a
function of s and t which is differentiable.
That is, z = f (g (s, t), h(s, t)) and
∂z ∂z ∂x ∂z ∂y ∂z ∂z ∂x ∂z ∂y
= + = +
∂s ∂x ∂s ∂y ∂s ∂t ∂x ∂t ∂y ∂t
∂z
= fx (g (s, t), h(s, t)) gs (s, t) + fy (g (s, t), h(s, t)) hs (s, t)
∂s
∂z
= fx (g (s, t), h(s, t)) gt (s, t) + fy (g (s, t), h(s, t)) ht (s, t)
∂t
MATH 127 (Section 11.5) The Chain Rule The University of Kansas 5 / 11
∂z ∂z
If z = x 2 y + 2xy 4 , x = st 2 , and y = s 2 t. Find and .
∂s ∂t
Solution:
∂z ∂z ∂x ∂z ∂y
= +
∂s ∂x ∂s ∂y ∂s
= (2xy + 2y 4 )(t 2 ) + (x 2 + 8xy 3 )(2st)
= (2s 3 t 3 + 2s 8 t 4 )(t 2 ) + (s 2 t 4 + 8s 7 t 5 )(2st)
∂z ∂z ∂x ∂z ∂y
= +
∂t ∂x ∂t ∂y ∂t
= (2xy + 2y 4 )(2st) + (x 2 + 8xy 3 )(s 2 )
= (2s 3 t 3 + 2s 8 t 4 )(2st) + (s 2 t 4 + 8s 7 t 5 )(t 2 )
In general, if z = f (x1 , x2 , . . . , xn ), x1 = g1 (t1 , t2 , . . . , tm ), . . . ,
xn = gn (t1 , t2 , . . . , tm ).
∂z ∂f ∂x1 ∂f ∂f ∂f ∂xn
= + + ... +
∂tj ∂x1 ∂tj ∂x2 ∂tj ∂xn ∂tj
MATH 127 (Section 11.5) The Chain Rule The University of Kansas 6 / 11
Implicit Differentiation
dy Fx (x, y )
If y = y (x) is defined implicitly by F (x, y ) = 0, then =− .
dx Fy (x, y )
Take the derivative with respect to x through F (x, y (x)) = 0 to get
dy
Fx (x, y )(1) + Fy (x, y ) = 0.
dx
dy
Example: Find if x 3 + y 3 − 6xy = 0.
dx
Solution: F (x, y ) = x 3 + y 3 − 6xy .
Fx (x, y ) = 3x 2 − 6y Fy (x, y ) = 3y 2 − 6x
dy Fx (x, y ) x 2 − 2y
Thus =− =− 2
dx Fy (x, y ) y − 2x
If z = z(x, y ) is defined implicitly by F (x, y , z) = 0 then
∂z Fx (x, y , z) ∂z Fy (x, y , z)
=− =−
∂x Fz (x, y , z) ∂y Fz (x, y , z)
MATH 127 (Section 11.5) The Chain Rule The University of Kansas 7 / 11
Suppose the equation implicitly defines a function xz 2 + y 2 z + xy = 1
where z depends upon x and y , z = z(x, y ). Find the partial derivatives
implicitly.
Solution: Let F (x, y , z) = xz 2 + y 2 z + xy − 1,
Fx (x, y , z) = z 2 +y Fy (x, y , z) = 2yz +x Fz (x, y , z) = 2zx +y 2
∂z Fx (x, y , z) z2 + y
=− =−
∂x Fz (x, y , z) 2zx + y 2
∂z Fy (x, y , z) 2yz + x
=− =−
∂y Fz (x, y , z) 2zx + y 2
MATH 127 (Section 11.5) The Chain Rule The University of Kansas 8 / 11
A baseball player hits the ball and then runs
down the first base at 20 ft/s. The first
baseman fields the ball and then runs towards
first base along the second base line at 18/, ft/s.
Determine how fast the distance between the
two-players is changing at a moment when the
hitter is 8 ft from first base and the baseman is
6 ft from first base.
The distance D between the runner, who is R ft from first, and the
baseman, who is B ft from first, is implicitly described by D 2 = R 2 + B 2 .
Both variables depend upon time t, B = B(t) and R = R(t).
∂D R ∂D B dB dR
= , = , = −18 , = −20
∂R D ∂B D dt dt
Using the chain rule,
dD ∂D dR ∂D dB 6 8 132
= + = (−20) + (−18) = − = −26.4 ft/s
dt ∂R dT ∂B dt 10 10 5
MATH 127 (Section 11.5) The Chain Rule The University of Kansas 9 / 11
