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 dzdt=∂z∂xdxdt+∂z∂ydydt.
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 dzdt=∂z∂xdxdt+∂z∂ydydt=(2xy)(2t)+(x2−2y)(2)=(2t2⋅2t)(2t)+((t2)2−2(2t))(2)=8t4+2t4−8t=10t4−8t.
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.
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)(√v2√u)+(x2ex2y)(0)=2√uv⋅1ve(√uv)2⋅1v⋅√v2√u+(√uv)2⋅e(√uv)2⋅1v⋅(0)=eu+0=eu∂z∂v=∂z∂x∂x∂v+∂z∂y∂y∂v=(2xyex2y)(√u2√v)+(x2ex2y)(−1v2)=2√uv⋅1ve(√uv)2⋅1v⋅√u2√v+(√uv)2e(√uv)2⋅1v⋅(−1v2)=uveu–uveu=0.
These Chain Rules generalize to functions of three or more variables in a straight forward manner.
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
dzdt=∂z∂xdxdt+∂z∂ydydt.
- 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.