How do you differentiate y = (3x - 5) / (3x^2 + 7)?

1 Answer
Jan 5, 2016

Use the quotient rule.

Answer:

f'(x)=(-9x^2+30x+21)/(3x^2+7)^2

Explanation:

Quotient rule

For function

f(x)=u/v

f'(x)=(vu'-uv')/v^2

So:

For f(x)=(3x-5)/(3x^2+7)

let
u=3x-5
so
u'=3
let
v=3x^2+7
so
v'=6x

f'(x)=((3x^2+7)*3-(3x-5)*6x)/(3x^2+7)^2

Expand and simplify:

f'(x)=(9x^2+21-18x^2+30x)/(3x^2+7)^2

f'(x)=(-9x^2+30x+21)/(3x^2+7)^2