The Chain Rule

In an introductory calculus course, it is shown that the chain rule is instrumental in both the theory and the applications of derivatives. In this section, we generalize the all-important chain rule to functions of two or more variables.

Let's suppose that w = f( x,y) is differentiable at (p,q) , and then let's suppose that x = h( t) , y = k(t) where h and k are differentiable at t₀ and (h( t₀) ,k( t₀) ) = ( p,q) . Then definition 3.2 in section 3 says that

Dw = fx( p,q) Dx+fy( p,q) Dy + o1(Dx,Dy)Dx + o2(Dx,Dy)Dx

so that if we divide by Dt and let Dt approach 0, then we obtain

lim Dt® 0   Dw Dt = fx( p,q) lim Dt® 0   Dx Dt +fy( p,q) lim Dt® 0   Dy Dt + lim Dt® 0  o1(Dx,Dy)  Dx Dt +o2(Dx,Dy) Dy Dt

(1)

In order for dx/dt to exist, it is necessary that Dx approaches 0 as Dt approaches 0. Likewise, Dy approaches 0 as Dt approaches 0. Since o₁(Dx,Dy) and o₂(Dx,Dy) approach 0 as Dx and Dy approach 0, the last limit in (1) becomes

lim Dt® 0  o1(Dx,Dy)  Dx Dt +o2(Dx,Dy) Dy Dt  =  lim Dt® 0  o1(Dx,Dy)  dx dt + lim Dt® 0  o2(Dx,Dy) dy dt  = 0

Consequently, (1) reduces to

dw dt = fx( p,q)  dx dt +fy( p,q)  dy dt

(2)

which is known as the chain rule for functions of 2 variables. Moreover, the chain rule can also be written in the form

dw dt =  ¶f ¶x  dx dt +  ¶f ¶y  dy dt

EXAMPLE 1    Find dw/dt given that w = x²+y³ and that x = cot( t) , y = sin( t) .       

Solution: The first partial derivatives of w = x²+y³ are

¶w ¶x = 2x,         ¶w ¶y = 3y2

As a result, the chain rule says that

dw dt = 2x  dx dt +3y2  dy dt

and since dx/dt = -csc²( t) and dy/dt = cos( t), we have

dw dt = 2cot( t) · [ -csc2(t) ] +3[ sin( t) ]2 cos(t) = -2cot( t) csc2( t) +3sin2(t) cos( t)

       

EXAMPLE 2    Find dw/dt given that w = cos( xy) and that x = pe^t and y = e^-t.   

Solution: The first partial derivatives are

¶w ¶x = -ysin( xy) ,         ¶w ¶y = -xsin( xy)

As a result, the chain rule says that

dw dt = -ysin( xy)  dx dt -xsin( xy)  dy dt

and since dx/dt = pe^t and dy/dt = -e^-t, we have

dw dt = -e-tsin( pete-t) ( pet) -petsin( pete-t) (-e-t) = -pete-tsin( pete-t) +pete-tsin( pete-t) = -psin( p) +psin( p) = 0