Multi-Variable Chain Rule

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The Multivariable Chain Rule – HMC Calculus Tutorial

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=zxdxdt+zydydt.

Since z=f(x,y) is differentiable at the point (x,y), Δz=zxΔx+zyΔy+ε1Δx+ε2Δy where ε10 and ε20 as (Δx,Δy)(0,0). Thus, ΔzΔt=zxΔxΔt+zyΔyΔt+ε1ΔxΔt+ε2ΔyΔt. Taking the limit as Δt0, limΔt0ΔzΔt=limΔt0[zxΔxΔt+zyΔyΔt+ε1ΔxΔt+ε2ΔyΔt]dzdt=zxdxdt+zydydt+(limΔt0ε1)dxdt+(limΔt0ε2)dydt. But limΔt0Δx=limΔt0ΔxΔtΔt=dxdtlimΔt0Δt=0 and similarly limΔt0Δy=0, so limΔt0ε1=limΔt0ε2=0. Thus, dxdt=xxdxdt+zydydt+(0)dxdt+(0)dydt=xxdxdt+zydydt. (proof taken from Calculus, by Howard Anton.)

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
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Let z=x2yy2 where x and y are parametrized as x=t2 and y=2t.

Then dzdt=zxdxdt+zydydt=(2xy)(2t)+(x22y)(2)=(2t22t)(2t)+((t2)22(2t))(2)=8t4+2t48t=10t48t.

Since x(t)=t2 and y(t)=2t, z=x2yy2=(t2)2(2t)(2t)2=2t54t2. We can now compute dzdt directly! dzdt=10t48t, as we obtained using the Chain Rule.

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 zu=zxxu+zyyuzv=zxxv+zyyv.

Holding v fixed and applying the first multivariable Chain Rule in this tutorial to z = z(x(u),y(u)), zu=zxxu+zyyu. Similarly, holding u fixed and applying the Chain Rule to z=z(x(u),y(u)), zv=zxxv+zyyv. (proof taken from Calculus, by Howard Anton.)

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

Example
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Let z=ex2y, where x(u,v)=uv and y(u,v)=1/v. Then zu=zxxu+zyyu=(2xyex2y)(v2u)+(x2ex2y)(0)=2uv1ve(uv)21vv2u+(uv)2e(uv)21v(0)=eu+0=euzv=zxxv+zyyv=(2xyex2y)(u2v)+(x2ex2y)(1v2)=2uv1ve(uv)21vu2v+(uv)2e(uv)21v(1v2)=uveuuveu=0.

Since x=uv and y=1v, z=ex2y=e(uv)2(1v)=eu. Then, differentiating z with respect to u and v, respectively, zu=euzv=0 as found using the Chain Rule!

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=zxdxdt+zydydt.

  • 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 zu=zxxu+zyyuzv=zxxv+zyyv.

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