| The Multivariable Chain Rule |
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
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dz/dt
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�z/
�x
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dx/dt
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�z/
�y
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dy/dt
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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
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�z/
�x
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dx/
dt
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�z/
�y
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dy/
dt
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| (2t2�2t)(2t)
+ ( (t2)2-2(2t) ) (2) |
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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
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�z/
�x
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�x/
�u
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+ |
�z/
� y
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�y/
�u
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�z/
�x
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�x/
�v
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+ |
�z/
� y
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�y/
�v
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Again, the variable-dependence diagram shown here indicates this Chain Rule by
summing paths for z either to u or to v.
Key Concepts
- 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
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dz/
dt
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= |
�z/
�x
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dx/
dt
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+ |
�z/
�y
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dy/
dt
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- 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
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�z/
�x
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�x/
�u
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+ |
�z/
� y
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�y/
�u
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�z/
�x
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�x/
�v
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+ |
�z/
� y
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�y/
�v
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