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

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Answer 1 · 2016-01-05T00:05:46

Answer:

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

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:

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

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

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