Mathematics
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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
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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.
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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
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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)
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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 :)
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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!