Looking for someone to check my answer:
Find an equation of the tangent line through the given point.
x 2 y 3 + 15y = 34x, (3, 2)

- anonymous

- jamiebookeater

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- anonymous

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

- anonymous

thats the answer i got.

- anonymous

well that's what i had for y'

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## More answers

- asnaseer

You first need to use implicit differentiation to find an expression for \(\displaystyle\frac{dy}{dx}\).
Then substitute x=3 and y=2 into that expression to get the slope of the tangent line at that point.
So then you will know the slope of the tangent line and you also know it passes through the point (3,2) - use this information to calculate the equation of the tangent line.

- anonymous

was my y' equation correct?

- asnaseer

it doesn't look correct to me - can you please list your steps so that I can help spot where you may have made a mistake?

- anonymous

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

- asnaseer

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

- anonymous

lol so it is.

- anonymous

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

- anonymous

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

- asnaseer

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

- asnaseer

you haven't used the chain rule correctly

- asnaseer

I mean "product rule"

- asnaseer

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

- anonymous

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

- asnaseer

that is correct :)

- anonymous

okay so now i isolate y'

- asnaseer

exactly

- anonymous

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

- asnaseer

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

- asnaseer

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

- anonymous

ok.

- asnaseer

the general rule of thumb is to write in this order:
1. Constants first
2. Then letters in alphabetical order

- anonymous

okay so now i sub in my points right?

- asnaseer

correct

- anonymous

so i had: -14/123

- asnaseer

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

- anonymous

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

- asnaseer

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

- asnaseer

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

- anonymous

so how would you make it look?

- asnaseer

ok, you got to this equation:\[y-2=-14/123x+42/123\]first add 2 to both sides to get:\[y=-14x/123 + 288/123\]then multiply both sides by 123 - what will you get then?

- anonymous

y123=-14x+288

- asnaseer

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

- anonymous

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

- asnaseer

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

- asnaseer

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

- asnaseer

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

- asnaseer

One last advice before I leave - you wrote one of the terms in your original equation as:
-14/123x
this can sometimes be confused for: \[-\frac{14}{123x}\]so it is usually better to write it as:
-14x/123

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