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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)

Mathematics
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\[\frac{ 34-2x }{ 2y+15 }\]
thats the answer i got.
well that's what i had for y'

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Other answers:

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.
was my y' equation correct?
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?
\[x ^{2}y ^{2}+15y=34x\] \[2x*2yy'+15y'=34\] \[y'(2y+15)=34-2x\] \[y'=\frac{ 34-2x }{ 2y+15 }\]
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.
the general rule of thumb is to write in this order: 1. Constants first 2. Then letters in alphabetical order
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?
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?
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: \(y=mx +c\)
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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