Implicit differentiation of 2sin(x)cos(y)=1

[asdfsystema]

Homework Statement

what is the implicit differentiation ?

Homework Equations

2sin(x)cos(y)=1

The Attempt at a Solution

d/dx[2sin(x)cos(y)]= d/dx[1]

2cos(x)*-sin(y)*dy/dx=0

I haev a bad feeling i did this wrong...

 

[Mark44]

To differentiate 2sin(x)cos(y), you have to use the product rule. You also need to use the chain rule when you differentiate a function of y (where y is assumed to be implicitly a function of x). l'm pretty sure you realized this, since you ended up with a factor of dy/dx, but you forgot the product rule.

After you have differentiated both sides, solve algebraically for dy/dx and you're done.

 

[asdfsystema]

I also graphed this using a program:

Are these appropriate ranges to find the implicit differentiation? I'm using this picture so it'll be easier for me to find out what the implicit differentiation of this is.

x and y ranges -6.28 to 6.28 ... are these appropriate figures or can I use a different value that shows it even better ? Thanks in advance.

Edit: Thanks Matt44, give me a sec !
Edit#2:
using the product rule as you suggested (finally makes sense...)
d/dx[2sin(x)cos(y)]=
u= 2sin(x) u'=2cos(x) v= cos(y) v'= -sin(y)*dy/dx
=== 2sin(x)*-sin(y)*dy/dx + cos(y) * 2cos(x)
=== Is the answer dy/dx= -cos(y)*2cos(x)/ -2sin(x)*sin(y) ?

Doesnt look too right ... ><

 

[HallsofIvy]

Cancelling "-2" in numerator and denominator gives dy/dx= cos(x)cos(y)/sin(x)sin(y) and that is exactly right.

 

[asdfsystema]

thanks hallofivy ! do you think -6.28 and 6.28 are good ranges for x and y ? or should it become something smaller ?

Matt44 and HallsOfIvy . u two are always there to answer my problems haha. thanks a lot :) !

 

[HallsofIvy]

What do you mean by "good ranges"? For a graph? Yes, those will do. The larger you make range the more "hills and valleys" you will see.

 

[asdfsystema]

sorry for the confusion. yup, that's what i meant . thank you.