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Harvey Mudd College Math Tutorial:
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, multi- plying derivatives along each path.
Example
Let z = x^2 y − y^2 where x and y are parametrized as x = t^2 and y = 2t. Then
dz dt
∂z ∂x
dx dt
∂z ∂y
dy dt = (2xy)(2t) + (x^2 − 2 y)(2) = (2t^2 · 2 t)(2t) +
( (t^2 )^2 − 2(2t)
) (2) = 8 t^4 + 2t^4 − 8 t = 10 t^4 − 8 t
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 par- tial 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 = ex (^2) y , where x(u, v) =
uv and y(u, v) = 1/v. Then
∂z ∂u
∂z ∂x
∂x ∂u
∂z ∂y
∂y ∂u =
( 2 xyex (^2) y)
v 2
u
)
( x^2 ex (^2) y) (0)
uv ·
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 =
( 2 xyex (^2) y)
u 2
v
)
( x^2 ex (^2) y) ( −
v^2
)
uv ·
v e(
√uv) (^2) · 1 v (^) ·
u 2
v
uv)^2 e(
√uv) (^2) · 1 v (^) ·
( −
v^2
)
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.
