Question

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

Answer 1

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)`