The Multivariate Chain Rule, Exams of Calculus

Download The Multivariate Chain Rule and more Exams Calculus in PDF only on Docsity!

The Multivariate Chain Rule

Ryan C. Daileda

Trinity University

Calculus III

Introduction

The chain rule in Calculus I tells us how to differentiate compositions: functions of the form f (g (x)).

A composition can be thought of as starting with a function f (t) and then replacing its variable with another function: t = g (x).

We can form compositions of functions of several variables in an analogous manner, by replacing the given variables with functions of new variables.

The process of finding the partial derivatives of such compositions is called the Chain Rule.

We directly substitute:

f (x, y ) = x^2 + y 3 = t^2 sin^2 s + (t^4 + s^2 )^3

and then compute:

∂f ∂t

= 2t sin^2 s + 3(t^4 + s^2 )^24 t^3 = 2t sin s

∂x ∂t

• 3(t^4 + s^2 )^2

∂y ∂t

= 2x

∂x ∂t

• 3y 2

∂y ∂t

∂f ∂x

∂x ∂t

∂f ∂y

∂y ∂t

∂f ∂s

= 2t^2 sin s cos s + 3(t^4 + s^2 )^22 s = 2t sin s

∂x ∂s

• 3(t^4 + s^2 )^2

∂y ∂s

= 2x

∂x ∂s

• 3y 2

∂y ∂s

∂f ∂x

∂x ∂s

∂f ∂y

∂y ∂s

Tree Diagrams

The chain rule explains the results of the preceding example.

We start by introducing a tree diagram:

Start with a root vertex labelled f.

Below f draw branches to vertices labelled with each of the original independent variables.

Below each of the each independent variables draw branches to each of the new independent variables.

We label each branch in a given tree diagram with the partial derivative of the vertex above with respect to the vertex below.

So in the second example above we get the labelling:

f

x y

r s

z

r s r^ s

∂f ∂x

∂f ∂z

∂x ∂r

∂x ∂s

∂y ∂r

∂y ∂s

∂z ∂r

∂z ∂s

∂f ∂y

To compute the derivative of f with respect to a variable in the bottom row, we follow every path to that variable, multiplying as we go, and add the results.

Continuing with our example, there are three paths (in red) to the variable r (in blue):

f

x y

r s

z

r s r^ s

∂f ∂x

∂f ∂z

∂x ∂r

∂x ∂s

∂y ∂r

∂y ∂s

∂z ∂r

∂z ∂s

∂f ∂y

“Multiplying down and adding across” gives the result

∂f ∂r

∂f ∂x

∂x ∂r

∂f ∂y

∂y ∂r

∂f ∂z

∂z ∂r

We now substitute x = r cos θ and y = r sin θ:

∂f ∂r

= 2xy 3 cos θ + 3x^2 y 2 sin θ

= 5r 4 cos^2 θ sin^3 θ ,

∂f ∂θ

= − 2 xy 3 r sin θ + 3x^2 y 2 r cos θ

= r 5 (−2 cos θ sin^4 θ + 3 cos^3 θ sin^2 θ).

Example 3

If w = xey^ /z^ and x = t^2 , y = 1 − t and z = 1 + 2t, compute

dw dt

Solution. We have the following tree diagram:

w

x y

∂w ∂x (^) ∂w ∂y z

∂w ∂z

dx dt

dy dt

dz dt t t^ t

dw dt

∂w ∂x

dx dt

∂w ∂y

dy dt

∂w ∂z

dz dt

= ey^ /z^ (2t) +

x z

ey^ /z^ (−1) −

xy z^2

ey^ /z^ (2)

= e

1 −t 1+2t

2 t −

t^2 1 + 2t

2 t^2 (1 − t) (1 + 2t)^2

= e

1 −t 1+2t

8 t^3 + 5t^2 + 2t (1 + 2t)^2

Thus

gu (u, v ) = fx (x, y )eu^ + fy (x, y )eu = fx (eu^ + sin v , eu^ + cos v )eu^ + fy (eu^ + sin v , eu^ + cos v )eu^ ,

gv (u, v ) = fx (x, y ) cos v − fy (x, y ) sin v = fx (eu^ + sin v , eu^ + cos v ) cos v − fy (eu^ + sin v , eu^ + cos v ) sin v ,

so that, according to the table,

gu (0, 0) = fx (e^0 + sin 0, e^0 + cos 0)e^0 + fy (e^0 + sin 0, e^0 + cos 0)e^0 = fx (1, 2) + fy (1, 2) = 2 + 5 = 7 ,

gu (0, 0) = fx (e^0 + sin 0, e^0 + cos 0) cos 0 − fy (e^0 + sin 0, e^0 + cos 0) sin 0 = fx (1, 2) = 2.

Example 5

Use the Multivariate Chain Rule to derive the ordinary Product and Quotient Rules from Calculus I.

Solution. Given f (x) and g (x), let P(u, v ) = uv and Q(u, v ) = u/v.

Then P(f (x), g (x)) = f (x)g (x) and Q(f (x), g (x)) = f g^ ((xx)).

The chain rule gives:

du

P

u

∂P ∂u

v

∂P ∂v

dv dx dx x x

d dx

(f (x)g (x)) =

dP dx

∂P

∂u

du dx

∂P

∂v

dv dx = v

df dx

• v

dg dx = g (x)

df dx

• f (x)

dg dx

Implicit Differentiation

Given a differentiable function f (x, y ), the Implicit Function Theorem states that the implicit curve with equation

f (x, y ) = 0

defines y as a differentiable function of x at every point where fy 6 = 0.

In Calculus I you learn the process of implicit differentiation for computing dy dx.

If we use the Chain Rule we can derive a much faster procedure involving the partial derivatives of f.

Setting y = y (x) and differentiating both sides of f (x, y ) = 0 with respect to x, the chain rule gives:

dx

f

x

∂f ∂x

y

∂f ∂y

dy dx dx x x

df dx

∂f ∂x

dx dx

∂f ∂y

dy dx

=

∂f ∂x

∂f ∂y

dy dx

Solving for dy dx we obtain

dy dx

−∂f /∂x ∂f /∂y

In a similar way, given a differentiable function F (x, y , z), the implicit surface defined by

F (x, y , z) = 0

defines z as a function of x and y wherever Fz 6 = 0.

Setting z = z(x, y ) and applying the Chain Rule we obtain

∂F

∂x

∂F

∂x

∂x ∂x

∂F

∂y

∂y ∂x

∂F

∂z

∂z ∂x

=

∂F

∂x

∂F

∂z

∂z ∂x

Hence ∂z ∂x

−∂F /∂x ∂F /∂z

Likewise one can show that

∂z ∂y

−∂F /∂y ∂F /∂z

Example 7

If x^2 y + y 2 z + z^2 x = xyz, find

∂z ∂x

and

∂z ∂y

Solution. The given equation is equivalent to

F (x, y , z) = x^2 y + y 2 z + z^2 x − xyz = 0.

Thus ∂z ∂x

−∂F /∂x ∂F /∂z

−(2xy + z^2 − yz) y 2 + 2xz − xy