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Mathematics 18 Online
OpenStudy (anonymous):

I got the right answer, but need to understand HOW to get to the answer.. help pls? Find the derivative of the function. h(x) = (5 + x)/(2x^2 +5)^2

OpenStudy (anonymous):

Do you know the rule you need for derivating a fractional function?

OpenStudy (anonymous):

yes, quotient rule.

OpenStudy (anonymous):

ok, here goes

OpenStudy (anonymous):

\[f'(x)g(x) - f(x)g'(x) \over { g(x)^2 } \]

OpenStudy (anonymous):

My simplified answer was correct, but it wasn't the completely simplified version.

OpenStudy (anonymous):

f'(x) = 1 g'(x) = 2(2x^2+5)*4x

OpenStudy (anonymous):

got that too?

OpenStudy (anonymous):

yep

OpenStudy (anonymous):

what about posting the answer you got and the answer you need?

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

OpenStudy (anonymous):

let me know if you can't view it

OpenStudy (anonymous):

i'm viewing looks like you need to cancel out x in all terms

OpenStudy (anonymous):

you factor out (2x^2+5) in all terms

OpenStudy (anonymous):

(5+x)(15x^3+40x) = -8 (x-5) x (2 x+5)

OpenStudy (anonymous):

-8 (x-5) (2x^2+5x)

OpenStudy (anonymous):

after that step all exp[ressions share the (2x^2+5x) factor and you can cancel them out

OpenStudy (anonymous):

brb

OpenStudy (anonymous):

I think you have a typo... it should be factored by (2x^2 +5)? no "x" with the 5

OpenStudy (anonymous):

at least that's how it is on the answer :)

OpenStudy (anonymous):

true, but that's the way to do it, make sure they have a common factor and get rid of it

OpenStudy (anonymous):

oh ok.. got it. thank you!

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