How do you find the derivative of # (3+sin(x))/(3x+cos(x))#?

1 Answer
benjidunn
May 20, 2016

Use quotient rule to obtain #(f/g)'=(3xcosx-8)/(9x^2+6xcosx+cos^2x)#

Explanation:

Quotient rule: #(f/g)'=(f'g-g'f)/g^2#
Let #f=3+sinx#; therefore #f'=cosx#
Let #g=3x+cosx#; therefore #g'=3-sinx#
#(f/g)'=(f'g-g'f)/g^2#
#=[cosx(3x+cosx)-(3-sinx)(3+sinx)]/(3x+cosx)^2#
#=[(3xcosx+cos^2x)-(9-sin^2x)]/(9x^2+6xcosx+cos^2x)#
#=(3xcosx+sin^2x+cos^2x-9)/(9x^2+6xcosx+cos^2x)#
Identity: #sin^2x+cos^2x=1=>(f/g)'=(3xcosx-8)/(9x^2+6xcosx+cos^2x)#

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