The Multivariable Chain Rule

Suppose that z = f(x,y), where x and y themselves depend on one or more variables. Multivariable Chain Rules allow us to differentiate z with respect to any of the variables involved:

Let x = x(t) and y = y(t) be differentiable at t and suppose that z = f(x,y) is differentiable at the point (x(t),y(t)). Then z = f(x(t),y(t)) is differentiable at t and
dz
dt
 = ∂z
∂x
 dx
dt
 + ∂z
∂y
 dy
dt
 .

Proof

Although the formal proof is not trivial, the variable-dependence diagram shown here provides a simple way to remember this Chain Rule. Simply add up the two paths starting at z and ending at t, multiplying derivatives along each path.

Example

Let z = x2y-y2 where x and y are parametrized as x = t2 and y = 2t.

Then

dz
dt
=
∂z
∂x
 dx
dt
 + ∂z
∂y
 dy
dt
=
(2xy)(2t) + (x2-2y)(2)
=
(2t2·2t)(2t) + ( (t2)2-2(2t) ) (2)
=
8t4 + 2t4 -8t
=
10t4-8t

Alternate Solution

We now suppose that x and y are both multivariable functions.

Let x = x(u,v) and y = y(u,v) have first-order partial derivatives at the point (u,v) and suppose that z = f(x,y) is differentiable at the point (x(u,v),y(u,v)). Then f(x(u,v),y(u,v)) has first-order partial derivatives at (u,v) given by
∂z
∂u
=
∂z
∂x
 ∂x
∂u
 + ∂z
∂ y
 ∂y
∂u
∂z
∂v
=
∂z
∂x
 ∂x
∂v
 + ∂z
∂ y
 ∂y
∂v

Proof

Again, the variable-dependence diagram shown here indicates this Chain Rule by summing paths for z either to u or to v.

Example

Let z = ex2y, where x(u,v) = √[uv] and y(u,v) = 1/v. Then

∂z
∂u
=
∂z
∂x
 ∂x
∂u
 + ∂z
∂y
 ∂y
∂u
=
( 2xyex2y )





√v
2√u





+ (x2ex2y)(0)
=
2   __
√uv
 
 · 1
v
 e(√[uv])2 · 1/v · √v
2√u
 + (   __
√uv
 
 )2 · e(√[uv])2 ·1/v ·(0)
=
eu + 0
=
eu
∂z
∂v
=
∂z
∂x
 ∂x
∂v
 + ∂z
∂y
 ∂y
∂v
=
( 2xyex2y )





√u
2√v





+ (x2ex2y)





- 1
v2





=
2   __
√uv
 
 · 1
v
 e(√[uv])2·1/v · √u
2√v
 + (   __
√uv
 
 )2e(√[uv])2 ·1/v ·





- 1
v2





=
u
v
 eu - u
v
 eu
=
0.

Alternate Solution

These Chain Rules generalize to functions of three or more variables in a straight forward manner.

Key Concepts [index]

  • Let x = x(t) and y = y(t) be differentiable at t and suppose that z = f(x,y) is differentiable at the point (x(t),y(t)). Then z = f(x(t),y(t)) is differentiable at t and
    dz
    dt
    		 = ∂z
    ∂x
    		 dx
    dt
    		 + ∂z
    ∂y
    		 dy
    dt
    		 .

  • Let x = x(u,v) and y = y(u,v) have first-order partial derivatives at the point (u,v) and suppose that z = f(x,y) is differentiable at the point (x(u,v),y(u,v)). Then f(x(u,v),y(u,v)) has first-order partial derivatives at (u,v) given by
    ∂z
    ∂u
    =
    ∂z
    ∂x
    		 ∂x
    ∂u
    		 + ∂z
    ∂ y
    		 ∂y
    ∂u
    ∂z
    ∂v
    =
    ∂z
    ∂x
    		 ∂x
    ∂v
    		 + ∂z
    ∂y
    		 ∂y
    ∂v