What is the derivative of #( cos (x) ) / ( 2 + sin (x) )#?

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Answer 1 · 2018-07-27T19:43:24

#f'(x)=\frac{-2\sin x-1}{(2+\sin x)^2}#

Given function:

#f(x)=\frac{\cos x}{2+\sin x}#

Differentiating above function w.r.t. #x# using quotient rule as follows

#d/dxf(x)=d/dx(\frac{\cos x}{2+\sin x})#

#f'(x)=\frac{(2+\sin x)d/dx(\cos x)-\cos xd/dx(2+\sin x)}{(2+\sin x)^2}#

#=\frac{(2+\sin x)(-\sin x)-\cos x(\cos x)}{(2+\sin x)^2}#

#=\frac{-2\sin x-\sin^2x-\cos ^2x}{(2+\sin x)^2}#

#=\frac{-2\sin x-(\sin^2x+\cos ^2x)}{(2+\sin x)^2}#

#=\frac{-2\sin x-1}{(2+\sin x)^2}#

Answer 2 · 2018-07-27T19:51:15

#-(2sinx+1)/(2+sinx)^2#

We have a quotient here, so we can find the derivative using quotient rule which, if you forgot, is:

#d/dx(g(x)/(h(x))) = (g'(x)*h(x) - h'(x)*g(x))/(h^2(x))#

So let's apply it:

#f(x)=cosx/(2+sinx)#

#f'(x)=\frac{(2+\sin x)d/dx(\cos x)-(\cos x)d/dx(2+\sin x)}{(2+\sin x)^2}#

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

#f'(x)=(-2sinx-sin^2x-cos^2x)/(2+sinx)^2#

#f'(x)=(-2sinx-(color(red)(sin^2x+cos^2x)))/(2+sinx)^2#

To simplify this further, we can use the pythagorean identity:
#color(red)(sin^2x+cos^2x=1#

So:

#f'(x)=(-2sinx-1)/(2+sinx)^2# or #-(2sinx+1)/(2+sinx)^2#