Calculus 3: Chain Rule for Finding dx/dy with x=yz and y=2sin(y+z)

[AlonsoMcLaren]

Homework Statement

If x=yz and y=2sin(y+z), find dx/dy

Homework Equations

Chain rule

The Attempt at a Solution

From y = 2sin(y+z) we get
dz/dy= (1-2cos(y+z))/(2cos(y+z))
dz/dy=((1/2)sec(y+z) - 1)

dx/dy = ∂x/∂y + ∂x/∂z dz/dy
= z + y ((1/2)sec(y+z) - 1)
= z - y - (1/2) sec(y+z)

But the answer is z-y+tan(y+z) (Mathematical methods in the physical sciences, Boas, Ch. 4 Sec 7 Problem 1)

What is going wrong?

 

[tiny-tim]

Hi AlonsoMcLaren!

From y = 2sin(y+z) we get
dz/dy= (1-2cos(y+z))/(2cos(y+z))

Nooo

 

[AlonsoMcLaren]

Hi AlonsoMcLaren!


Nooo

So how to calculate dz/dy?

 

[tiny-tim]

one line at a time, for a start!

show us your first line ​

 

[AlonsoMcLaren]

y - 2sin(y+z) = 0

Let f=y - 2sin(y+z)
Then ∂f/∂y = 1 - 2cos(y+z)
∂f/∂z = -2cos(y+z)

0 = df = (∂f/∂y)dy+(∂f/∂z)dz
= (1-2cos(y+z)) dy - 2 cos(y+z) dz

dz/dy = (1-2cos(y+z))/(2cos(y+z))

 

[dimension10]

Homework Statement

If x=yz and y=2sin(y+z), find dx/dy

Homework Equations

Chain rule

The Attempt at a Solution

From y = 2sin(y+z) we get
dz/dy= (1-2cos(y+z))/(2cos(y+z))
dz/dy=((1/2)sec(y+z) - 1)

dx/dy = ∂x/∂y + ∂x/∂z dz/dy
= z + y ((1/2)sec(y+z) - 1)
= z - y - (1/2) sec(y+z)

But the answer is z-y+tan(y+z) (Mathematical methods in the physical sciences, Boas, Ch. 4 Sec 7 Problem 1)

What is going wrong?

First of all differentiate y=2sin(y+z) with respect to y. From chain rule, we obtain

$$ 1=2cos(y+z) $$

Now solve for y and then I think you can solve it.

 

[tiny-tim]

oops!

sorry, that is right

you've just left out a y when you expanded this bracket …

dx/dy = ∂x/∂y + ∂x/∂z dz/dy
= z + y ((1/2)sec(y+z) - 1)
= z - y - (1/2) sec(y+z)

 

[AlonsoMcLaren]

sorry, that is right

you've just left out a y when you expanded this bracket …

So how does it lead to z-y+tan(y+z)?

 

[tiny-tim]

x/dy = ∂x/∂y + ∂x/∂z dz/dy

= z + y ((1/2)sec(y+z) - 1)

= z - y - (1/2) y sec(y+z)

= … ? ​