What is the derivative of #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=u/v# for #u# and #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)#