How do you find the derivative of $sinx/(1+cosx)$ ?

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Answer 1 · 2018-05-03T17:24:19

$1/(cosx+1)$

$f(x)=sinx/(cosx+1)$

$f'(x)=(sinx/(cosx+1))'$

The derivative of $f(x)/g(x)$ using Quotient Rule is
$(f'(x)g(x)-f(x)g'(x))/g^2(x)$
so in our case it is

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

$(cosx(cosx+1)+sin^2x)/(cosx+1)^2$ $=$

$(color(blue)(cos^2x)+cosx+color(blue)(sin^2x))/(cosx+1)^2$ $=$

$cancel((cosx+color(blue)(1)))/(cosx+1)^cancel(2)$ $=$

$1/(cosx+1)$

Answer 2 · 2018-05-03T18:08:38

$1/2sec^2(x/2) or 1/(1+cosx)$ .

We have, $sinx/(1+cosx)$ ,

$={2sin(x/2)cos(x/2)}/{2cos^2(x/2)}$ ,

$=tan(x/2)$ .

$"Therefore, "d/dx{sinx/(1+cosx)}$ ,

$=d/dx{tan(x/2)}$ ,

$=sec^2(x/2)*d/dx{x/2}......"[The Chain Rule]"$ ,

$=sec^2(x/2)*1/2$ ,

$=1/2sec^2(x/2), or,$

$=1/(2cos^2(x/2))$ ,

$=1/(1+cosx)$ .

How do you find the derivative of sinx/(1+cosx) sinx/(1+cosx)?