anonymous
  • anonymous
3) Differentiate using the quotient rule. g(x) = (x^2+5x)/(3x^2+4) The quotient rule f(x) / g(x) = g(x) * f ’ (x) – f(x) * g ’(x) / (g(x)^2 ) (3x^2 + 4)
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
@amistre64
amistre64
  • amistre64
theres not much to do except for apply the quotient rule that was defined.
anonymous
  • anonymous
yes the part i am confused about is the derivative of the function

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

amistre64
  • amistre64
it unfortunate that the definition uses the same notation g(x) that the question uses
amistre64
  • amistre64
\[(\frac{t}{b})'=\frac{bt'-b't}{b^2}\] define the top and bottom, along with their rather simple derivates .. they are just polynomials after all
amistre64
  • amistre64
whats is the top function? what is its derivative?
anonymous
  • anonymous
for example (x^2+5x) would be 2x + 5? its been awhile since I used these rules
amistre64
  • amistre64
good, lets fill that in: \[(\frac{t}{b})'=\frac{bt'-b't}{b^2}\] \[(\frac{x^2+5x}{b})'=\frac{b(2x+5)-b'(x^2+5x)}{b^2}\] what is the bottom function? and its square? what is its derivative?
amistre64
  • amistre64
ideally the square is just ^2 ... and usually doenst require the expansion
anonymous
  • anonymous
(3x^2+42x+1-x^2+5x23x)/(3x^2+4)2
amistre64
  • amistre64
it looks like you tried to jump to the end with that
amistre64
  • amistre64
when learning the basics, its usually a good idea to step thru it to get the feel of it down pat
anonymous
  • anonymous
yes I agree
amistre64
  • amistre64
b = 3x^2+4 b' = 6x filling in we have: \[(\frac{x^2+5x}{b})'=\frac{b(2x+5)-b'(x^2+5x)}{b^2} \] \[(\frac{x^2+5x}{3x^2+4})'=\frac{(3x^2+4)(2x+5)-(6x)(x^2+5x)}{(3x^2+4)^2} \] the top is usually expanded afterwards and simplified .... but this is usually deemed good as is
anonymous
  • anonymous
so you apply the rule
amistre64
  • amistre64
thats is the rule applied
anonymous
  • anonymous
isnt it suppose to be the derivative of g(x) and f(x)
anonymous
  • anonymous
okay nvm lol
anonymous
  • anonymous
I see it now 6x
amistre64
  • amistre64
no, it is spose to be the derivative of the top and bottom function .. what you name them is totally irrelevant
anonymous
  • anonymous
Its all coming back to me now lol I just needed a refresher.
amistre64
  • amistre64
:) good luck
anonymous
  • anonymous
Okay so now what would be the next step distributing or canceling like terms?
amistre64
  • amistre64
the next step, if needed, would be to use the stuff from algebra to simplify the results to your hearts content; yes
anonymous
  • anonymous
Also i have a few others like this if you don't mind helping me get back up to speed
amistre64
  • amistre64
i can see what i can do ....

Looking for something else?

Not the answer you are looking for? Search for more explanations.