How do you find the first and second derivatives of $(3x-2)^2/(e^(2x)-5)$ using the quotient rule?

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Answer 1 · 2018-05-31T12:22:00

$f'(x)=(2(2x-5)(5e^(2x)-3xe^(2x)-15))/(e^(2x)-5)^2$

We Need the Quotient rule
$f'(x)=(u'v-uv')/v^2$
with
$u=(3x-2)^2$
so
$u'=2(3x-2)3$
$v=e^(2x)-5$
$v'=2e^(2x)$
and we get

$f'(x)=(2(3x-2)*3*(e^(2x)-5)-(3x-2)^2*e^(2x)*2)/(e^(2x)-5)^2$
This simplifies to

$(2(3x-5)*(3e^(2x)-15-3e^(2x)x+2e^(2x)))/(e^(2x)-5)^2$
so
$f'(x)=(2(3x-5)(5e^(2x)-3xe^(2x)-15))/(e^(2x)-5)^2$

How do you find the first and second derivatives of (3x-2)^2/(e^(2x)-5)(3x-2)^2/(e^(2x)-5) using the quotient rule?