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

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Answer 1 · 2016-10-18T19:16:38

$f'(x) = {(x)cos(x)}/(x + cos(x))^2$

Given: $f(x) = g(x)/(h(x))$

$f'(x) = {g'(x)h(x) - g(x)h'(x)}/(h(x))^2$

$g(x) = 1 + sin(x)$
$g'(x) = cos(x)$
$h(x) = x + cos(x)$
$h'(x) = 1 - sin(x)$

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

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

$f'(x) = {(x)cos(x) + cos^2(x) - (cos^2(x))}/(x + cos(x))^2$

$f'(x) = {(x)cos(x)}/(x + cos(x))^2$

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