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

2 Answers
May 9, 2017

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

Explanation:

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)

May 9, 2017

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

:. f'(x)=-1/(1+sinx).

Explanation:

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.!