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Chain rule

CHAIN RULE FOR MULTIVARIABLE FUNCTIONS

1. If x = e

t

, y = t

2

− 3 , and z = 2 t + 1 , use chain rule to find df / dt.

f ( x , y , z ) = xy

2

z

3

Solution:

Find derivatives of the parametric curve.

dx

dt

= e

t

dy

dt

= 2 t

dz

dt

Find partial derivatives of f ( x , y , z ).

∂ f

∂ x

= y

2

z

3

∂ f

∂ y

= 2 xyz

3

∂ f

∂ z

= 3 xy

2

z

2

Use chain rule to find df / dt.

and

∂ θ

∂ ϕ

∂ θ

∂ ϕ

π

Find partial derivatives of z at ϕ = π / 4.

∂ z

∂ r

= 2 r sin θ

∂ z

∂ r

= 2 ϕ

2

sin (

ϕ + π )

∂ z

∂ r

π

π

2

sin

π

• π

∂ z

∂ r

π

π

2

∂ z

∂ r

π

π

2

and

∂ z

∂ θ

= r

2

cos θ

∂ z

∂ θ

= ϕ

4

cos (

ϕ + π )

∂ z

∂ θ

π

π

4

cos

π

• π

∂ z

∂ θ

π

π

4

∂ z

∂ θ

π

π

4

Use chain rule to find dz / dϕ.

dz

dϕ

∂ z

∂ r

dr

dϕ

∂ z

∂ θ

dθ

dϕ

dz

dϕ

π

2

π

π

4

dz

dϕ

π

3

π

4

1. If u = ln( 3 t ) and v = ln t with t > 0 , use chain rule to find the global

maximum of the function.

f ( u , v ) = 3 u − 2 v

2

Solution:

Find derivatives of u and v.

t = e

3 / 4

Since f ′( t ) > 0 for t < e

3 / 4

, and f ′( t ) < 0 for t > e

3 / 4

, then t = e

3 / 4

is the global

maximum of the function.

f ( e

3 / 4

) = 3 ln( 3 e

3 / 4

) − 2 (ln( e

3 / 4

2

f ( e

3 / 4

) = ln( 27 ) +

2

2

f ( e

3 / 4

• ln( 27 )

CHAIN RULE FOR MULTIVARIABLE FUNCTIONS AND TREE DIAGRAMS

1. If x = sin( t + s ), y = 2 ts , and z = 2 t − 5 s , use chain rule to find the partial

derivatives f

t

and f

s

f ( x , y , z ) = 7 x + 2 y

2

z

Solution:

Find partial derivatives with respect to s and t of x , y , and z.

x

s

= cos( t + s ) x

t

= cos( t + s )

y

s

= 2 t y

t

= 2 s

z s

= − 5 z t

Find partial derivatives for f ( x , y , z ).

∂ f

∂ x

∂ f

∂ y

= 4 yz

∂ f

∂ z

= 2 y

2

Use chain rule to find ∂ f /∂ t.

1. If x = log

2

( ts ) and y = log

3

( 2 t + s ), use chain rule to find partial derivatives

f

s

and f

t

at (1,1).

f ( x , y ) = x

2

− 2 xy − y

2

• x + 3 y − 4

Solution:

Evaluate x and y at (1,1).

x (1,1) = log 2

y (1,1) = log 3

Find partial derivatives of x and y at (1,1).

x

s

s ln( 2 )

ln( 2 )

x t

t ln( 2 )

ln( 2 )

and

y

s

( 2 t + s )ln( 3 )

3 ln( 3 )

y

t

( 2 t + s )ln( 3 )

3 ln( 3 )

Find partial derivatives f with respect to x and y.

∂ f

∂ x

= 2 x − 2 y + 1 = 2 ( 0 ) − 2 ( 1 ) + 1 = − 1

∂ f

∂ y

= − 2 x − 2 y + 3 = − 2 ( 0 ) − 2 ( 1 ) + 3 = 1

Use chain rule to find ∂ f /∂ s and ∂ f /∂ t at (1,1).

∂ f

∂ s

∂ f

∂ x

⋅ x s

∂ f

∂ y

⋅ y s

∂ f

∂ s

ln( 2 )

3 ln( 3 )

∂ f

∂ s

3 ln( 3 )

ln( 2 )

and

∂ f

∂ t

∂ f

∂ x

⋅ x t

∂ f

∂ y

⋅ y t

∂ f

∂ t

ln( 2 )

3 ln( 3 )

∂ f

∂ t

3 ln( 3 )

ln( 2 )

The partial derivatives of f at (1,1) with respect to s and t are

∂ f

∂ s

3 ln( 3 )

ln( 2 )

∂ f

∂ t

3 ln( 3 )

ln( 2 )

∂ f

∂ s

= − 10 ( 2 t − s ) + 7 ( t + 2 s ) + 2

∂ f

∂ s

= 24 s − 13 t + 2

and

∂ f

∂ t

∂ f

∂ x

⋅ x t

∂ f

∂ y

⋅ y t

∂ f

∂ t

= ( 4 x − 3 y )( 2 ) + (− 3 x + 2 y + 1 )( 1 )

∂ f

∂ t

= 5 x − 4 y + 1

∂ f

∂ t

= 5 ( 2 t − s ) − 4 ( t + 2 s ) + 1

∂ f

∂ t

= − 13 s + 6 t + 1

Solve the system of equations.

− 13 s + 6 t + 1 = 0

24 s − 13 t + 2 = 0

We get

− 13 s + 6 t = − 1

24 s − 13 t = − 2

then

− 169 s + 78 t = − 13

144 s − 78 t = − 12

Add the equations.

− 169 s + 78 t + ( 144 s − 78 t ) = − 13 + (− 12 )

− 169 s + 78 t + 144 s − 78 t = − 13 − 12

− 25 s = − 25

s = 1

Then

− 13 ( 1 ) + 6 t = − 1

6 t = 12

t = 2

So the point we’re looking for is ( s , t ) = (1,2).