The Chain Rule in Vector Form

To begin with, suppose that C is a curve in the xy-plane with a smooth parameterization x( t) = á x(t), y(t) ñ , t in [ a,b] , and suppose that f(x) is differentiable at each point on C. In the last part of the previous section, we saw that Definition 5.1b implies that on some neighborhood of 0, there is a continuous function e( Dx) with e( 0) = 0 such that

| Df - Ñf·Dx |  <  e( Dx) ||Dx||

If we now suppose that Dx = x( t+Dt)-x( t) for some small nonzero change Dt in the parameter, then

ê ê    Df Dt  - Ñf·  Dx  Dt ê ê < e( Dx) ||  Dx  Dt ||

If Dt approaches 0, then Dx also approaches 0 and so

lim Dt® 0    ê ê    Df Dt  - Ñf ·  Dx  Dt ê ê £ lim Dt® 0    æ è e( Dx) || Dx Dt || ö ø

which leads to

ê ê    df dt  - Ñf·  dx  dt ê ê  £ e( 0) || dx  dt ||

Since e(0) = 0, this implies that following:  

      

The Chain Rule:  If x( t) is differentiable at each point t in an interval [ a,b] , and if f( x) is differentiable at each x(t) , t in [ a,b] , then

df dt   = Ñf · dx dt

      

Alternatively, the chain rule can be written as

d dt f( r) = Ñf · v

(1)

where v = x' (t) is the velocity vector of x( t) .  Also, if we let df/dx denote the gradient Ñf (i.e., df/dx = Ñf ) then the chain rule can be writtten in the form

df dt =  df dx · dx dt

which is reminiscent of the chain rule for functions of a single variable.        

EXAMPLE 1    Evaluate df/dt using (1) given that f( x,y) = x²-y² and x(t) = á sin(t), cos(t) ñ

Solution: Since Ñf = á 2x, -2y ñ and

v = x' (t) = á cos(t), -sin(t) ñ

the chain rule (1) implies that

df dt   = Ñf · v = á 2x,-2y ñ · á cos( t),-sin( t) ñ = 2xcos( t) +2ysin(t)

However, x( t) = á sin(t), cos(t) ñ implies that x = sin(t) and y = cos(t) , so that

df dt   = 2sin(t) cos( t) +2cos(t) sin( t) = 4sin(t) cos( t) = 2sin(2t)

   

Similarly, if w = U( x,y,z) and r(t) = á x(t), y(t), z(t) ñ is a curve in R³, then we let w = U( r) and let

ÑU =     ¶U ¶x ,  ¶U ¶y ,  ¶U ¶z

be the gradient of U, so that the chain rule for 3 variables can be written

dU dt = ÑU· dr dt

EXAMPLE 2    Evaluate dU/dt using (1) given that

U( x,y,z) = xy+z2

and r( t) = á e^t,e^-t,t³ ñ.        

Solution: Since ÑU = á y,x,2z ñ and r'  = á e^t,-e^-t,3t² ñ , the chain rule says that

dU dt   = ÑU ·    dr dt   = á y,x,2z ñ · áet,-e-t,3t2 ñ = yet-xe-t+6zt2

Since x = e^t, y = e^-t, and z = t³, we have

dU dt = e-tet-ete-t+6t3t2 = 6t5

Check your Reading: How many trig identities did we use in example 1?