How do you find the derivative of G(x) = (4-cos(x))/(4+cos(x))G(x)=4−cos(x)4+cos(x)?

1 Answer
Mia
Sep 12, 2016

(8sinx)/(4+cosx)^28sinx(4+cosx)2

Explanation:

The derivative of the quotient is defined as follows:
(u/v)'=(u'v-v'u)/v^2

Let u=4-cosx and v=4+cosx

Knowing that color(blue)((d(cosx))/dx=-sinx)

Let us find u' and v'

u'=(4-cosx)'=0-color(blue)((-sinx))=sinx
v'=(4+cosx)'=0+color(blue)((-sinx))=-sinx

G'(x)=(u'v-v'u)/v^2
G'(x)= (sinx(4+cosx)-(-sinx)(4-cosx))/(4+cosx)^2
G'(x)=(4sinx+sinxcosx+4sinx-sinxcosx)/(4+cosx)^2
G'(x)=(8sinx)/(4+cosx)^2

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