How do you differentiate #f(x)=cosx/(1+sinx)#?

Question

Charles Anctil · Accepted Answer

#:. f'(x)=secxtanx-sec^2x=secx(tanx-secx), or, #

#:. f'(x)=-1/(1+sinx).# #f(x)=cosx/(1+sinx)=cosx/(1+sinx)*(1-sinx)/(1-sinx).# #=(cosx(1-sinx))/(1-sin^2x)=(cosx(1-sinx))/cos^2x# #=(1-sinx)/cosx=1/cosx-sinx/cosx# #:. f(x)=secx-tanx.# #:. f'(x)=secxtanx-sec^2x=secx(tanx-secx), or, # #f'(x)=secxtanx-sec^2x=1/cosx*sinx/cosx-1/cos^2x, # #=(sinx-1)/cos^2x=(sinx-1)/((1+sinx)(1-sinx))# #:. f'(x)=-1/(1+sinx).# Enjoy Maths.!

William Abdalla · Answer

#(df)/(dx)=-1/(1+sinx)#

We use quotient rule. Let #g(x)=cosx# and #h(x)=1+sinx#

then #(dg)/(dx)=-sinx# and #(dh)/(dx)=cosx#

and hence using quotient rule

#(df)/(dx)=((1+sinx)xx(-sinx)-cosx xx cosx)/(1+sinx)^2#

= #(-sinx-sin^2x-cos^2x)/(1+sinx)^2#

= #-(1+sinx)/(1+sinx)^2#

= #-1/(1+sinx)#

Aurora Alvarado · Answer

undefined To differentiate $ f(x) = \frac{\cos(x)}{1+\sin(x)} $, you can use the quotient rule, which states that if you have a function of the form $ \frac{u(x)}{v(x)} $, then its derivative is given by $ \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $.

Applying the quotient rule to the given function, you'll have:

$ u(x) = \cos(x) $ and $ v(x) = 1 + \sin(x) $.

$ u'(x) = -\sin(x) $ and $ v'(x) = \cos(x) $.

Substitute these into the quotient rule formula:

$ f'(x) = \frac{(-\sin(x))(1+\sin(x)) - \cos(x)(\cos(x))}{(1+\sin(x))^2} $.

Simplify the expression:

$ f'(x) = \frac{-\sin(x) - \sin^2(x) - \cos^2(x)}{(1+\sin(x))^2} $.

Using the trigonometric identity $ \sin^2(x) + \cos^2(x) = 1 $:

$ f'(x) = \frac{-\sin(x) - 1}{(1+\sin(x))^2} $.