The Chain Rule

Review Of The Chain Rule For One Variable

Recall that if

y  =  f(x)      and      x  =  x(t)

then

dy            dy       dx
=
dt             dx        dt

Example

Suppose that

z  =  f(x,y)  =  x2 + 2x - xy + y2

and

x(t)  =  t2 +1            y(t)  =  t3 - t2

Then what is dz/dt for t = 2?

Solution:

f can be written as

[x(t)]2 + 2[x(t)] - x(t)y(t) + [y(t)]2

Hence the derivative is

2x(t)[x'(t)] + 2(2t) - [x(t)y'(t) + y(t)x'(t)] + 2y(t)y'(t)

=  2x(t)[2t] + 2(2t) - [x(t)(3t2 - 2t) + y(t)(2t)] + 2y(t)(3t2 - 2t)

Now since t = 2

x(2)  =  5       and       y(2)  =  4

We can substitute in to get

2(5)[2(2)] + 2(2)(2) - [(5)(3(2)2- 2(2)) + (4)2(2))] + 2(4)(3(2)2- 2(2))

=  56

Alternatively we use the chain rule:

 A Chain Rule With this chain rule the derivative becomes

(2x + 2 - y)(2t) + (-x + 2y)(3t2 - 2t)

When

t = 2,      x(2) = 5

and

y(2)  =  8 - 4  =  4

hence

df / dt  =  [2(5) + 2 - 4](2)(2) + [-5 + 2(4)] [3(4) - 2(2)]  =  56

Exercise:

Let

f(x,y) = 2x -3xy

and

x(q)  =  2 cos q      and      y(q)  = 2 sin q

Find df/dq

The Chain Rule for 2 Variables
 A Chain Rule For Two Variables Let f(x,y) be a 2 variable function and            x = x(u,v)      and      y = y(u,v)  then Example: Polar Coordinates

Let

f(x,y)  =  x2

and

x  =  r cos q     and     y  =  r sin q

then =  2xy cos q + x2 sin q

=  2r2 cos2 q sin q + r2cos2 q sin q

=  3r2cos2q sinq

Exercise

Let

f(x,y)  =  x - 2y2

and

x(u,v)  =  u - 2v        y(u,v)  =  2u + v

find Implicit Differentiation

Suppose we have the ellipsoid

x2 + y2  + 2z2  =  1

and we want to find We write

f(x,y,z)  =  f(x, y, z(x,y))  =  0

but Where the first factors correspond to the partials with respect to the first, second and third variables respectively and the second factors are with respect to the actual x.  We have

Jx/Jx = 1     and        Jy/Jx = 0
Hence

0 = Jf/Jx + Jf/Jz Jz/Jx

So that

Jz/Jx = -Jf/Jx / Jf/Jz = -fx/fz

for the ellipsoid:

Jf/Jx = 2x/4z = x/2z

In general,
 zx = -fx/fz    and     zy = -fy/fz

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