y'=1/1/sinx i have no idea were to start because the 1/1/sinx is throwing me off big time
M matt757 New member Joined Mar 24, 2015 Messages 38 May 5, 2015 #1 y'=1/1/sinx i have no idea were to start because the 1/1/sinx is throwing me off big time
D Deleted member 4993 Guest May 5, 2015 #2 matt757 said: y'=1/1/sinx i have no idea were to start because the 1/1/sinx is throwing me off big time Click to expand... If your problem looks like this: \(\displaystyle \displaystyle{y' \ = \ \frac{1}{\frac{1}{sin(x)}}}\) then \(\displaystyle \displaystyle{ \frac{1}{\frac{1}{sin(x)}} = sin(x)}\) and continue....
matt757 said: y'=1/1/sinx i have no idea were to start because the 1/1/sinx is throwing me off big time Click to expand... If your problem looks like this: \(\displaystyle \displaystyle{y' \ = \ \frac{1}{\frac{1}{sin(x)}}}\) then \(\displaystyle \displaystyle{ \frac{1}{\frac{1}{sin(x)}} = sin(x)}\) and continue....
Steven G Elite Member Joined Dec 30, 2014 Messages 14,364 May 5, 2015 #3 matt757 said: y'=1/1/sinx i have no idea were to start because the 1/1/sinx is throwing me off big time Click to expand... a/b/c=(a/b)/c. So 1/1/sinx = (1/1)/sinx = 1/sinx
matt757 said: y'=1/1/sinx i have no idea were to start because the 1/1/sinx is throwing me off big time Click to expand... a/b/c=(a/b)/c. So 1/1/sinx = (1/1)/sinx = 1/sinx