trying to find the second derivative in this question......3x-3 over 2x+4. the Vertical Asy is -2 and the Horizontal Asy is 3/2...the xint is (1,0) and the yint is (0,-3/4) please give me step buy step

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.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

trying to find the second derivative in this question......3x-3 over 2x+4. the Vertical Asy is -2 and the Horizontal Asy is 3/2...the xint is (1,0) and the yint is (0,-3/4) please give me step buy step

Mathematics
See more answers at brainly.com
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.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

can you find the first derivative? because that is the only hard part
quotient rule give \[\frac{d}{dx}\frac{3x-3}{2x+4}=\frac{(2x+4)\times 3-(3x+3)\times 2}{(2x+4)^2}\]
yes i have the first one all ready

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

so you end up with \[\frac{9}{(2x+4)^2}=9(2x+4)^{-2}\] now second should be easy
\[-18(2x+4)^{-3}\times 2=-36(2x+4)^{-3}=-\frac{36}{(2x+4)^3}\]
you can cancel a little because \[(2x+4)^3=(2(x+2))^3=8(x+2)^3\]
im so confused
ok which step? did you get the first derivative?
i got (2x+4)-2(3x-3) over (2x+4)^2
then you have to do the algebra in the numerator
should be \[(2x+4)3-2(3x-3) \]
in front of the 2x+4 there is suppose to be a 3
yes it is the denominator times the derivative of the numerator as the first term
minus the numerator times the derivative of the denominator
if you multiply out the x terms add up to zero and you get 9
so 18 over (2x+4)^2 ?
yes
\[(2x+4)3-2(3x-3)\] \[6x+12-6x+6=18\] yes
my equation is 3(2x+4)-2(3x-3)
yes multiply out and you will get 18
ok then what do i do to get the second answer
you have \[\frac{18}{(2x+4)^2}\] so rather than using the quotient rule again, rewrite in exponential form as \[18\times (2x+4)^{-2}\] and use the power rule (and the chain rule)
ok so multiple it out?
power rule right? \[\frac{d}{dx}x^n=nx^{n-1}\]
i made a mistake, sorry
\[\frac{d}{dx}18(2x+4)^{-2}=-2\times 18(2x+4)^{-2-1}\times 2\] \[=-72(2x+4)^{-3}=-\frac{72}{(2x+4)^3}\] but you can simplify this
because \[(2x+4)^3=(2(x+2))^3=8(x+2)^3\] and \[\frac{72}{8}=9\]
so your "final answer" will be \[-\frac{9}{(x+2)^3}\]
thank you so much
yw

Not the answer you are looking for?

Search for more explanations.

Ask your own question