Posted by Mary on Thursday, March 7, 2013 at 11:38am.
Find the derivative of sin(2x + 3y)= 3xy + 5y2.
I have this so far
1) cos(2x+3y) * d/dx (2x+3y)= 3xy' + 3y + 5y'
2) cos (2x+3y) * [2 + 3y']= 3xy' + 3y + 5y'
3) 3y' cos (2x+3y) + 2cos(2x+3y)= 3xy' + 3y + 5y'
4) 3xy' + 5y'  (3y(cos(3x+3y)= 2 cos (2x + 3y)  3y
If I am wrong, can you please explain to me what I am doing wrong. Thank you

Derive  Steve, Thursday, March 7, 2013 at 12:02pm
Aside from the typos in (4) you are ok.
4) 3xy' + 5y'  3cos(2x+3y)y' = 2 cos (2x + 3y)  3y
5) (3x+5y3cos(2x+3y))y' = 2cos(2x+3y)  3y
6) y' = (2cos(2x+3y)3y)/(3x+5y3cos(2x+3y))

Derive  Mary, Thursday, March 7, 2013 at 12:13pm
Thank you SO much sir. I feel a lot better knowing that I did this right
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