## mathcalculus one year ago help please: Use implicit differentiation to find the equation of the tangent line to the curve at the point. (attached below)

1. mathcalculus

2. mathcalculus

i know i have to use implicit differentiation, then plug into derivative to find slope, plug slope and given points to point-slope formula.

3. mathcalculus

Problem I have is step 1

4. phi

take the derivative with respect to x. use the product rule for example, on the xy term you do this $\frac{d}{dx}xy = x \frac{d}{dx}y+ y \frac{d}{dx}x \\ = x \frac{dy}{dx}+ y \frac{dx}{dx} \\ = x \frac{dy}{dx}+ y$ use the same idea on $$xy^3$$

5. mathcalculus

i got x(3y*dy/dx) +y^3

6. mathcalculus

@phi

7. mathcalculus

so.. 3xy*dydx +y^3

8. mathcalculus

dy/dx**

9. phi

you mean 3 x y^2 dy/dx for the first part d/dx of y^3 is 3 y^2 dy/dx

10. mathcalculus

okay im lost with the second part.

11. mathcalculus

i know that you took the derivative of xy^3= 3xy^2

12. mathcalculus

+xy= 1?

13. mathcalculus

*dy/dx

14. phi

from the beginning $\frac{d}{dx}\left(xy^3 + xy = 4\right) \\ x \frac{d}{dx}y^3 + y^3 \frac{d}{dx}x+ x\frac{d}{dx}y+ y \frac{d}{dx}x=\frac{d}{dx}4$ that simplifies to $x \frac{d}{dx}y^3 + y^3+ x\frac{dy}{dx}+ y =0$ now you can do the d/dx of y^3 , right ?

15. mathcalculus

yes.

16. phi

use the power rule $\frac{d y^n}{dx} = n y^{n-1} \frac{dy}{dx}$

17. mathcalculus

this isn't applying chain rule or is it? because i understand what youre doing here makes sense/

18. phi

yes, it is the chain rule

19. phi

normally you do d/dx of x^3 and you get 3x^2 dx/dx but the dx/dx is left off and the rule looks like $\frac{d\ x^3}{dx} = 3 x^2$

20. mathcalculus

okay, let me give this a try.

21. mathcalculus

okay I got this down: x*3y^2(dy/dx) +3y +x(dy/dx)+y=0

22. phi

you mean y^3 not 3y , right ?

23. mathcalculus

oh so only find the derivative of [y^3]...

24. mathcalculus

thanks btw! :)

25. mathcalculus

then i should just plot the points for x and y

26. mathcalculus

wait a min. what should i do with dy/dx?

27. phi

solve for dy/dx and then sub in for x and y to get the value of dy/dx at the point (2,1)

28. mathcalculus

how?

29. phi

use algebra

30. mathcalculus

question: is it possible to divide this: -4x/-2y ????

31. mathcalculus

how the hell can someone divide this and get 2x/y?????

32. mathcalculus

33. mathcalculus

@asnaseer

34. phi

what part is confusing ? you have $3 x y^2\frac{dy}{dx} + y^3+ x\frac{dy}{dx}+ y =0$ plug in x=2 and y =1 and solve for dy/dx (which is the slope)

35. phi

at the point (2,1)

36. mathcalculus

no no, i know that. i was just looking at an example and i see that they divide -4x/-2y and got 2x/y

37. mathcalculus

it frustrated me to see this and not know how they even possibly do it.

38. phi

are you asking how to get from $\frac{-4x}{-2y} = \frac{2x}{y}$ ? -4/-2 is +2

39. mathcalculus

yes but with diferet variables?? 0.o

40. mathcalculus

different*

41. phi

what is the question ?

42. mathcalculus

oh the slope is -1/4! =]

43. phi

yes. also, the tangent line goes through point (2,1) so you can find the intercept.

44. mathcalculus

cool, how do you know this?

45. phi
46. mathcalculus

okie doke

47. mathcalculus

thanks @phi you helped me so much! ^_^ so thankful

48. phi

they asked you to find the tangent line through point (2,1) you found its slope = -1/4 you know it goes through 2,1 so in point-slope form you have y- 1 = (-1/4)(x-2) which you can change to slope-intercept form

49. mathcalculus

y=-1/4x+3/2

50. phi

yes