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.

See more answers at brainly.com

\[\frac{ 34-2x }{ 2y+15 }\]

thats the answer i got.

well that's what i had for y'

was my y' equation correct?

\[x ^{2}y ^{2}+15y=34x\]
\[2x*2yy'+15y'=34\]
\[y'(2y+15)=34-2x\]
\[y'=\frac{ 34-2x }{ 2y+15 }\]

I thought you had \(y^3\) in the equation listed in your question?

lol so it is.

\[y'=\frac{ 34-2x }{ 3y ^{2}+15 }\]

did i have it right the second time (right above what you just wrote)

sorry I meant:\[\frac{d}{dx}(x^2y^3)=(x^2)\frac{d}{dx}(y^3)+y^3\frac{d}{dx}(x^2)\]

you haven't used the chain rule correctly

I mean "product rule"

have a look here: http://www.1728.org/chainrul.htm

\[x ^{2}*3y ^{2}y'+y ^{3}2x+15y'=34\]

that is correct :)

okay so now i isolate y'

exactly

\[y'=\frac{ 34-2xy ^{3} }{ x ^{2}3y ^{2}+15 }\]

yup - now follow the other steps that I had listed above.

NOTE: we don't usually write an expression as \(x^23y^2\) - it is better to write it as \(3x^2y^2\)

ok.

okay so now i sub in my points right?

correct

so i had: -14/123

perfect! just a couple of more steps to go now :)

is the equation: \[y-2=-14/123x+42/123\]

yes - that looks correct. I wouldn't have separated the two constants here (the -2 and the 42/123)

you may also want to multiply both sides by 123 to remove the fractions from the final equation.

so how would you make it look?

y123=-14x+288

correct - but again, remember to write constants first - so 123y instead of y123

that's not really how you write the equation of a line though.

so I would write the final equation as:\[123y=288-14x\]

it is still an equation of a line. you can write it in "standard form" as follows:\[14x+123y=288\]

maybe you are only used to seeing it in the form: \(y=mx +c\)