## Calculus 3: Chain rule

1. The problem statement, all variables and given/known data

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

2. Relevant equations

Chain rule

3. 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?

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Hi AlonsoMcLaren!
 Quote by AlonsoMcLaren From y = 2sin(y+z) we get dz/dy= (1-2cos(y+z))/(2cos(y+z))
Nooo

 Quote by tiny-tim Hi AlonsoMcLaren! Nooo
So how to calculate dz/dy?

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## Calculus 3: Chain rule

one line at a time, for a start!
 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))

 Quote by AlonsoMcLaren 1. The problem statement, all variables and given/known data If x=yz and y=2sin(y+z), find dx/dy 2. Relevant equations Chain rule 3. 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.

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