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

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Answer 1 · 2017-05-09T17:08:19

#(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)#

Answer 2 · 2017-05-09T17:08:30

#:. 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.!