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)

Question

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 · Answer

$$\frac{ 34-2x }{ 2y+15 }$$

anonymous · Answer

thats the answer i got.

anonymous · Answer

well that's what i had for y'

asnaseer · Answer

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 · Answer

was my y' equation correct?

asnaseer · Answer

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 · Answer

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

asnaseer · Answer

I thought you had $y^3$ in the equation listed in your question?

anonymous · Answer

lol so it is.

anonymous · Answer

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

anonymous · Answer

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

asnaseer · Answer

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

asnaseer · Answer

you haven't used the chain rule correctly

asnaseer · Answer

I mean "product rule"

asnaseer · Answer

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

anonymous · Answer

$$x ^{2}*3y ^{2}y'+y ^{3}2x+15y'=34$$

asnaseer · Answer

that is correct :)

anonymous · Answer

okay so now i isolate y'

asnaseer · Answer

exactly

anonymous · Answer

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

asnaseer · Answer

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

asnaseer · Answer

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

anonymous · Answer

ok.

asnaseer · Answer

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

anonymous · Answer

okay so now i sub in my points right?

asnaseer · Answer

correct

anonymous · Answer

so i had: -14/123

asnaseer · Answer

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

anonymous · Answer

is the equation: $$y-2=-14/123x+42/123$$

asnaseer · Answer

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

asnaseer · Answer

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

anonymous · Answer

so how would you make it look?

asnaseer · Answer

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 · Answer

y123=-14x+288

asnaseer · Answer

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

anonymous · Answer

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

asnaseer · Answer

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

asnaseer · Answer

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

asnaseer · Answer

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

asnaseer · Answer

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