How do you determine the derivative of #y = ((cosx)/(1 -sinx))^2#?
1 Answer
Dec 15, 2016
Explanation:
We have:
#y= ((cosx)/(1 - sinx))^2#
#y = (cos^2x)/(1 - 2sinx + sin^2x)#
#y = (1 - sin^2x)/(1 - sinx)^2#
#y = ((1 + sinx)(1 - sinx))/((1 - sinx)(1 - sinx))#
#y = (1 +sinx)/(1 -sinx)#
By the quotient rule:
#y'= (cosx(1 -sinx) - (-cosx(1 +sinx)))/(1 -sinx)^2#
#y' = (cosx- cosxsinx + cosx + cosxsinx)/(1 - sinx)^2#
#y' = (2cosx)/(1 -sinx)^2#
Hopefully this helps!
