What is the derivative of $\frac{sin x}{1 + cos x}$ $sinx/(1+cosx)$ ?

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Answer 1 · 2015-06-03T21:25:19

Use the division's derivative formula:

For a given function g: $g = \frac{u}{v}$ $g=u/v$ for $u$ $u$ and $v \neq 0$ $v!=0$ other functions, the derivative of g is found as;

$g'=(u'v-uv')/v^2$

If we apply it to our case:

$f'(x)=((sinx)'(1+cosx)-sinx(1+cosx)')/(1+cosx)^2= (cosx(1+cosx)+sinxsinx)/(1+cosx)^2=(cosx+cos^2x+sin^2x)/(1+cosx)^2$

Using now the trigonometric rule: $cos^2x+sin^2x=1$ we finally get:

$f'(x)=(1+cosx)/(1+cosx)^2=1/(1+cosx)$

What is the derivative of sinx/(1+cosx)sinx1+cosxsinx/(1+cosx)?