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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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
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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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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
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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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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
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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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( 2xyex2y ) |
![](../../om/simboli/graffa1.gif)
![](../../om/simboli/graffa0.gif)
![](../../om/simboli/graffa0.gif)
![](../../om/simboli/graffa0.gif)
![](../../om/simboli/graffa0.gif)
![](../../om/simboli/graffa0.gif)
![](../../om/simboli/graffa4.gif) |
√v 2√u
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![](../../om/simboli/graffa3.gif)
![](../../om/simboli/graffa0.gif)
![](../../om/simboli/graffa0.gif)
![](../../om/simboli/graffa0.gif)
![](../../om/simboli/graffa0.gif)
![](../../om/simboli/graffa0.gif)
![](../../om/simboli/graffa2.gif) |
+ (x2ex2y)(0) |
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2 |
| __ √uv
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· |
1 v
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e(√[uv])2 · 1/v · |
√v 2√u
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+ ( |
| __ √uv
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)2 · e(√[uv])2 ·1/v ·(0) |
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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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( 2xyex2y ) |
![](../../om/simboli/graffa1.gif)
![](../../om/simboli/graffa0.gif)
![](../../om/simboli/graffa0.gif)
![](../../om/simboli/graffa0.gif)
![](../../om/simboli/graffa0.gif)
![](../../om/simboli/graffa0.gif)
![](../../om/simboli/graffa4.gif) |
√u 2√v
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![](../../om/simboli/graffa3.gif)
![](../../om/simboli/graffa0.gif)
![](../../om/simboli/graffa0.gif)
![](../../om/simboli/graffa0.gif)
![](../../om/simboli/graffa0.gif)
![](../../om/simboli/graffa0.gif)
![](../../om/simboli/graffa2.gif) |
+ (x2ex2y) |
![](../../om/simboli/graffa1.gif)
![](../../om/simboli/graffa0.gif)
![](../../om/simboli/graffa0.gif)
![](../../om/simboli/graffa0.gif)
![](../../om/simboli/graffa0.gif)
![](../../om/simboli/graffa0.gif)
![](../../om/simboli/graffa4.gif) |
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1 v2
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![](../../om/simboli/graffa3.gif)
![](../../om/simboli/graffa0.gif)
![](../../om/simboli/graffa0.gif)
![](../../om/simboli/graffa0.gif)
![](../../om/simboli/graffa0.gif)
![](../../om/simboli/graffa0.gif)
![](../../om/simboli/graffa2.gif) |
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2 |
| __ √uv
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· |
1 v
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e(√[uv])2·1/v · |
√u 2√v
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+ ( |
| __ √uv
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)2e(√[uv])2 ·1/v · |
![](../../om/simboli/graffa1.gif)
![](../../om/simboli/graffa0.gif)
![](../../om/simboli/graffa0.gif)
![](../../om/simboli/graffa0.gif)
![](../../om/simboli/graffa0.gif)
![](../../om/simboli/graffa0.gif)
![](../../om/simboli/graffa4.gif) |
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1 v2
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![](../../om/simboli/graffa3.gif)
![](../../om/simboli/graffa0.gif)
![](../../om/simboli/graffa0.gif)
![](../../om/simboli/graffa0.gif)
![](../../om/simboli/graffa0.gif)
![](../../om/simboli/graffa0.gif)
![](../../om/simboli/graffa2.gif) |
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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
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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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. |
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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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