Question

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)

Answer 1

@amistre64

Answer 2

theres not much to do except for apply the quotient rule that was defined.

Answer 3

yes the part i am confused about is the derivative of the function

Answer 4

it unfortunate that the definition uses the same notation g(x) that the question uses

Answer 5

\[(\frac{t}{b})'=\frac{bt'-b't}{b^2}\]
define the top and bottom, along with their rather simple derivates .. they are just polynomials after all

Answer 6

whats is the top function?
what is its derivative?

Answer 7

for example (x^2+5x) would be 2x + 5?
its been awhile since I used these rules

Answer 8

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?

Answer 9

ideally the square is just ^2 ... and usually doenst require the expansion

Answer 10

(3x^2+42x+1-x^2+5x23x)/(3x^2+4)2

Answer 11

it looks like you tried to jump to the end with that

Answer 12

when learning the basics, its usually a good idea to step thru it to get the feel of it down pat

Answer 13

yes I agree

Answer 14

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

Answer 15

so you apply the rule

Answer 16

thats is the rule applied

Answer 17

isnt it suppose to be the derivative of g(x) and f(x)

Answer 18

okay nvm lol

Answer 19

I see it now 6x

Answer 20

no, it is spose to be the derivative of the top and bottom function .. what you name them is totally irrelevant

Answer 21

Its all coming back to me now lol I just needed a refresher.

Answer 22

:) good luck

Answer 23

Okay so now what would be the next step distributing or canceling like terms?

Answer 24

the next step, if needed, would be to use the stuff from algebra to simplify the results to your hearts content; yes

Answer 25

Also i have a few others like this if you don't mind helping me get back up to speed

Answer 26

i can see what i can do ....