The Chain Rule

Chain Rule – Case I (One Independent Variable)Let be a differentiable function in its domain where and are both differentiable function of Then is a differentiable function of and

The Chain Rule Case I

The function is differentiable at if the partial derivatives and e and are continuous at

dydz z dx z

dt x dt y dt

zx

z

x y

t t

• start with z• z is a function of x and y• x and y are functions of t

zy

dxdt

dydt

• Put the appropriate derivatives along the branches

Tree diagram

• Multiply along each path and sum the contributions of both paths.

Chain Rule - Example 1Let where and . Find when

dydz z dx z

dt x dt y dt(4 )( 3sin ) ( 2 )(3cos ) x t y t

zx

dxdt

zy

dydt

W and therefore

Geometric interpretation: The parametric equationsdescribe a circle C in the - plane. Imagine walking on the surface while staying directly above the circle C.

is the rate of change of as C is traversed

Chain Rule - Example 2The power in a DC (direct current) component is Watts. If is increasing at (ohm) per minute and is decreasing at (Volt) per minute, find when and .

dP P dE P dRdt E dt R dt

2

22

E dE E dRR dt dtR

E is decreasing at 0.1 V/min:

0.5

2

22(110) 1100.1 0.575 75

R is increasing at 0.5 / min :

1.369 W/min

When and , the power is decreasing at 1.369 Watts per minute.

Chain Rule – Case II (Two Independent Variables)Let be a differentiable function of and where and are differentiable function of and . Then

The Chain Rule Case II

yz z x z

s x s y s

zx

z

x y

t t

• Put the appropriate derivatives along the branches

Tree diagram

• For multiply along each path ending in and sum the contributions of each branch.

s

• x and y are functions of and

zy

xs

ys

yz z x z

t x t y t

xt

s

yt

• For multiply along each path ending in and sum the contributions of each branch.

Chain Rule - Example 3Let where and . Find and

z z r zs r s s

2 22sin

2( cos )( ) ( )

r r ss t

e t ezr

rs

z

s2 2

sincos

r sts t

e

z z r zt r t t

2 22sin

2( cos )( ) ( )

r r ts t

e s ezr

rt

zt

2 2sincos

r tss t

e

z

rs

s

rt

s

t

zr

s t t

r 𝜃

z

Example 4: Let be a differentiable function of and where and z are differentiable function of , and . Derive formulas for and

Chain Rule - More general caseWe can generalize the chain rule to any number of variables.

yw x w w z

x s y s zws s

zr

xt

s

zs

wx

s t t

x 𝑧

rr

xs

xr

w

y

s t r

wy

wz

yr

ys

yt

zt

yw x w w z

x t y t zwt t

yw x w w z

x r y r zwr r

Chain Rule - Implicit Differentiation 2DAssume the equation defines implicitly as a differentiable function of that is, where for all in the domain of .

Assume and and are continuous. Differentiate with respect to .

( , ) 0 F x yx xChain rule: 0 x y

dydxF Fdx dx

0 x ydyF F dx

Solve for : xy

FdyFdx

Chain Rule - Example 5Consider the equation (a) Find

Here

2 3

4 26 205 6

xy

dy xy xFFdx y x y

(b) Find the equation of the tangent line to the curve at the point (1,1).

First note that the point is indeed on the curve since the equation is satisfied when substituting and . The slope of the tangent line is given by at the point

Substituting and yields

Equation of tangent line: or

Chain Rule - Implicit Differentiation 3DAssume the equation defines implicitly as a differentiable function of and that is, where for all and in the domain of .

Assume and , and are continuous.

Differentiate with respect to using the chain rule:

0 zx yyx zF F Fx x x

1 0 0 zx yzF F F x

Solve for

xz

FzFx

0 zxzF F x

Differentiate with respect to using the chain rule:

0 zx yyx zF F Fy y y

0 1 0 zx yzF F F y

0 zyzF F y

Solve for

yz

FzFy

Chain Rule - Example 6Find and if

Here

2 3

2 2

1

3

xz

y zx yzFzyFx xy zx yz

3

2 2

2

3

yz

z xyzF x yzzyFy xy zx yz