How do you differentiate y=(2+sinx)/(x+cosx)?

Question

I'm assuming I use f(x)/g(x)=((f'(x)g(x)-(f(x)g(x)))/[g(x)]^2), but I can't seem to get it to work for me. How exactly would I do something like this?

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Answer 1 · 2018-03-06T05:34:26

dy/dx = #(xcos(x) + sin(x) - 1)/(x + cos(x))^2#

# "First, let's recall the Quotient Rule:" #

# \qquad \qquad \qquad \qquad \qquad [ f(x) / g(x) ]^' \ = \ { g(x) f'(x) - f(x) g'(x) } / { g(x)^2 } \quad. #

# "We are given the function to differentiate:" #

# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad y \ = \ { 2 + sinx } / { x + cosx } \quad.#

Use the quotient rule to derive the following:

y' = #{ [ (x + cosx) ( 2 + sinx )' ] - [ ( 2 + sinx ) ( x + cosx )' ] } / ( x + cosx )^2#

y' = #{ [ (x + cosx) ( cosx ) ] - [ ( 2 + sinx ) ( 1 -sinx )] } / ( x + cos x)^2#

multiplying the numerator out gets you this:

y' = #{xcosx + cos^2x - ( 2 - 2 sinx + sinx - sin^2x ) }/(x + cos)^2#

#\quad \ \ # = #{xcosx + cos^2x - ( 2 - sinx - sin^2x ) }/(x + cos)^2#

#\quad \ \ # = #{xcosx + cos^2x - 2 + sinx + sin^2x }/(x + cos)^2#

#\quad \ \ # = #{xcosx + sinx - 2 + ( sin^2x + cos^2x ) }/(x + cosx)^2#

then the only simplification you can use is the trig identity

#sin^2 + cos^2 = 1#

to get:

y' = #{xcosx + sinx - 2 + 1 }/(x + cosx)^2#

y' = #(xcos(x) + sin(x) - 1)/(x + cos(x))^2#