## bmelyk 3 years ago 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)

1. bmelyk

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

2. bmelyk

3. bmelyk

well that's what i had for y'

4. 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.

5. bmelyk

was my y' equation correct?

6. 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?

7. bmelyk

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

8. asnaseer

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

9. bmelyk

lol so it is.

10. bmelyk

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

11. bmelyk

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

12. asnaseer

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

13. asnaseer

you haven't used the chain rule correctly

14. asnaseer

I mean "product rule"

15. asnaseer

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

16. bmelyk

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

17. asnaseer

that is correct :)

18. bmelyk

okay so now i isolate y'

19. asnaseer

exactly

20. bmelyk

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

21. asnaseer

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

22. asnaseer

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

23. bmelyk

ok.

24. asnaseer

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

25. bmelyk

okay so now i sub in my points right?

26. asnaseer

correct

27. bmelyk

28. asnaseer

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

29. bmelyk

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

30. asnaseer

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

31. asnaseer

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

32. bmelyk

so how would you make it look?

33. 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?

34. bmelyk

y123=-14x+288

35. asnaseer

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

36. bmelyk

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

37. asnaseer

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

38. asnaseer

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

39. asnaseer

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

40. 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