How do you use the chain rule to differentiate $y=(cosx/(1+sinx))^5$ ?

Question

1408 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-09-20T19:23:00

$-5(cos^4x)/(1+sinx)^5$

$y=(cosx/(1+sinx))^5$
$dy/dx=d/dx((cosx/(1+sinx))^5)$
$=5(cosx/(1+sinx))^4*(((1+sinx)*(d/dx(cosx))-cosxd/dx(1+sinx))/(1+sinx)^2)$

$=5(cosx/(1+sinx))^4*(((1+sinx)*(-sinx)-cosx(cosx))/(1+sinx)^2)$

$=5(cosx/(1+sinx))^4*((-sinx-sin^2x-cos^2x)/(1+sinx)^2)$

$=5(cosx/(1+sinx))^4*((-sinx-(sin^2x+cos^2x))/(1+sinx)^2)$

$=5(cosx/(1+sinx))^4*((-sinx-1)/(1+sinx)^2)$

$=5(cosx/(1+sinx))^4*(-1)/(1+sinx)$
$=-5(cos^4x)/(1+sinx)^5$

How do you use the chain rule to differentiate y=(cosx/(1+sinx))^5y=(cosx/(1+sinx))^5?