find dy/dx sin^2 y=y+2

Question

this is an implicit differiation problem 

Answer 1

Hi Corey.

 

Everytime you take a derivative of a term with "y" in it, you must multiply the derivative by (dy/dx) because of the chain rule.

 

(siny)²=y+2

 

Power rule on left:  2(siny)(cosy)(dy/dx)

 

Right will be (dy/dx)+0

 

So you have 2(siny)(cosy)(dy/dx)=(dy/dx)

 

Put all the "(dy/dx)"'s on the left

 

2(siny)(cosy)(dy/dx)-(dy/dx)=0

 

Factor out a (dy/dx) and get

 

(dy/dx)[2sinycosy-1)=0

 

So either (dy/dx)=0  or  2sinycosy=0

 

Using trig identities, 2sinycosy = sin(2y)  so you can find the "y" that makes sin(2y)=0.

 

Sorry can't finish.  Lunch break is over...