Multivariable Chain Rule of sin(x)cos(2y)

[SweetBabyLou]

Hi all, I've got a Calculus III Question

Homework Statement

Find the derivative z_s and z_t, where z=sin(x)cos(2y)

Homework Equations

x=s+t
y=s-t

The Attempt at a Solution

I had a go at the solution and this was what I ended up getting

for z_s, I ended up getting (cosxcos2y)(1)-2sinxsin2y(1)

where the 1 at the end of the δx and δy were the partial derivatives of x=s+t and y=s-t

I subbed in s and t values for x and y, and I ended up with

z_s=((cos(s)+cos(t))(cos(2s)-cos(2t)))-((2sin(s)+2sin(t))(sin(2s)-sin(2t)))

I did the same process for t and got

z_t=((-cos(s)-cos(t))(-cos(2s)+cos(2t)))+((2sin(s)+2sin(t))(sin(2s)-sin(2t)))

this seems like a really unnecessarily long answer and I'm pretty sure I messed something up. I can't seem to find the mistake though (I have a feeling its right under my nose).

Also, I don't know if this is really the easiest way to go about doing these problems. I felt like the past few chapters we've learned in lecture have been kinda rushed. If there is any way where these problems could be solved in an easier way, that advice would be MUCH appreciated. Thanks

--my brain feels like its turning into mush

P.S. if there are any post-editing mistakes, please forgive me. I am not used to posting on this site yet.

 

[SammyS]

Hi all, I've got a Calculus III Question

Homework Statement

Find the derivative z_s and z_t, where z=sin(x)cos(2y)

Homework Equations

x=s+t
y=s-t

The Attempt at a Solution

I had a go at the solution and this was what I ended up getting

for z_s, I ended up getting (cosxcos2y)(1)-2sinxsin2y(1)

Hello SweetBabyLou. Welcome to PF !

You're correct up to this point.

where the 1 at the end of the δx and δy were the partial derivatives of x=s+t and y=s-t

What you have next, is incorrect.
cos(a+b) ≠ cos(a) + cos(b), etc.

Use angle addition identities.

I subbed in s and t values for x and y, and I ended up with

z_s=((cos(s)+cos(t))(cos(2s)-cos(2t)))-((2sin(s)+2sin(t))(sin(2s)-sin(2t)))

I did the same process for t and got

z_t=((-cos(s)-cos(t))(-cos(2s)+cos(2t)))+((2sin(s)+2sin(t))(sin(2s)-sin(2t)))

this seems like a really unnecessarily long answer and I'm pretty sure I messed something up. I can't seem to find the mistake though (I have a feeling its right under my nose).

Also, I don't know if this is really the easiest way to go about doing these problems. I felt like the past few chapters we've learned in lecture have been kinda rushed. If there is any way where these problems could be solved in an easier way, that advice would be MUCH appreciated. Thanks

--my brain feels like its turning into mush

P.S. if there are any post-editing mistakes, please forgive me. I am not used to posting on this site yet.

 

[SweetBabyLou]

Hi SammyS,

I'm sorry, but I'm not really familiar with Angle Addition Identities (It is probably something I've learned, but has slipped my mind). I did, however, look it up on the friendly neighborhood Google, and saw in Wolfram Alpha's MathWorld, that if I had sin(a+b) (or in this case, s+t) then the result should be something along the lines of, sin(a)cos(b)+sin(b)cos(a). I also see that for cos(a+b) the result should be cos(a)cos(b)-sin(a)sin(b). Am I headed in the right direction?

 

[SammyS]

Hi SammyS,

I'm sorry, but I'm not really familiar with Angle Addition Identities (It is probably something I've learned, but has slipped my mind). I did, however, look it up on the friendly neighborhood Google, and saw in Wolfram Alpha's MathWorld, that if I had sin(a+b) (or in this case, s+t) then the result should be something along the lines of, sin(a)cos(b)+sin(b)cos(a). I also see that for cos(a+b) the result should be cos(a)cos(b)-sin(a)sin(b). Am I headed in the right direction?

Yes, those are the angle addition identities .