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

Question

8543 views around the world

You can reuse this answer
Creative Commons License

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.!

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