Vector Form of the Chain Rule   

The chain rule is the sum of 2 products, and thus it can be written as the dot product of 2 vectors. That is,

dw dt =  ¶f ¶x  dx dt +  ¶f ¶y  dy dt =  ¶f ¶x ,  ¶f ¶y ·  dx dt ,  dy dt

(3)

Let's introduce some new notation to exploit this interpretation of the inner product.

If r( t) = á x( t) ,y(t) ñ is a curve in the xy-plane, then we let w = f( r) denote the function w( t) = f( x( t) ,y( t) ) . Moreover, the last vector in (3) is the velocity v of r( t) , which we denote as

v =  dr dt =  dx dt ,  dy dt

Next, we define the gradient of a function f( x,y) to be

Ñf =  ¶f ¶x ,  ¶f ¶y

where ``Ñ" is pronounced ``del'' and is the gradient operator. In terms of this new notation, (3) can be written as

d dt f( r) = Ñf·v

(4)

Other notations for the gradient of a function include grad(f) and df/dr. This last notation is especially nice because it allows us to write the chain rule (4) as

df dt =  df dr ·  dr dt

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

EXAMPLE 3    Evaluate df/dt using (4) given that f( x,y) = 2x²y+y² and r( t) = á cos( t) ,sin²( t) ñ       

Solution: Since Ñf = á 4xy,2x²+2y ñ , substitution of r( t) = á cos(t) ,sin²( t) ñ yields

Ñf( r) = á 4cos( t) sin2( t) ,2cos2( t) +2sin2( t) ñ = á 2sin( t) 2sin( t) cos(t) ,2 ñ = á 2sin( t) sin( 2t) ,2 ñ

Since v = á -sin( t) ,2sin( t)cos( t) ñ = á -sin( t) ,sin( 2t) ñ , the chain rule (4) implies that

df dt = Ñf·v = á 2sin( t)sin( 2t) ,2 ñ · á -sin(t) ,sin( 2t) ñ

Evaluating the inner product then leads to

df dt = -2sin2( t) sin( 2t) +2sin( 2t) = 2sin( 2t) ( 1-sin2( t) ) = 2sin( 2t) cos2( t)

       

EXAMPLE 4    Evaluate df/dt using (4) given that f( x,y) = sin( pxy) and r(t) = á e^t,e^-t ñ      

Solution: Since Ñf = á pycos( pxy) ,pxcos( pxy) ñ , substitution of r( t) = á e^t,e^-t ñ yields

Ñf( r) = á pe-tcos( pete-t) ,petcos( pete-t) ñ = á pe-tcos( p) ,petcos( p) ñ = á -pe-t,-pet ñ

Since v = á e^t,-e^-t ñ , the chain rule (4) implies that

df dt = Ñf( r) ·v = á -pe-t,-pet ñ · áet,-e-t ñ = -pete-t+pete-t = 0