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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)

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  1. mathcalculus
    • one year ago
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  2. mathcalculus
    • one year ago
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    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
    • one year ago
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    Problem I have is step 1

  4. phi
    • one year ago
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    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
    • one year ago
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    i got x(3y*dy/dx) +y^3

  6. mathcalculus
    • one year ago
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    @phi

  7. mathcalculus
    • one year ago
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    so.. 3xy*dydx +y^3

  8. mathcalculus
    • one year ago
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    dy/dx**

  9. phi
    • one year ago
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    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
    • one year ago
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    okay im lost with the second part.

  11. mathcalculus
    • one year ago
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    i know that you took the derivative of xy^3= 3xy^2

  12. mathcalculus
    • one year ago
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    +xy= 1?

  13. mathcalculus
    • one year ago
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    *dy/dx

  14. phi
    • one year ago
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    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
    • one year ago
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    yes.

  16. phi
    • one year ago
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    use the power rule \[ \frac{d y^n}{dx} = n y^{n-1} \frac{dy}{dx} \]

  17. mathcalculus
    • one year ago
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    this isn't applying chain rule or is it? because i understand what youre doing here makes sense/

  18. phi
    • one year ago
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    yes, it is the chain rule

  19. phi
    • one year ago
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    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
    • one year ago
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    okay, let me give this a try.

  21. mathcalculus
    • one year ago
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    okay I got this down: x*3y^2(dy/dx) +3y +x(dy/dx)+y=0

  22. phi
    • one year ago
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    you mean y^3 not 3y , right ?

  23. mathcalculus
    • one year ago
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    oh so only find the derivative of [y^3]...

  24. mathcalculus
    • one year ago
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    thanks btw! :)

  25. mathcalculus
    • one year ago
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    then i should just plot the points for x and y

  26. mathcalculus
    • one year ago
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    wait a min. what should i do with dy/dx?

  27. phi
    • one year ago
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    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
    • one year ago
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    how?

  29. phi
    • one year ago
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    use algebra

  30. mathcalculus
    • one year ago
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    question: is it possible to divide this: -4x/-2y ????

  31. mathcalculus
    • one year ago
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    how the hell can someone divide this and get 2x/y?????

  32. mathcalculus
    • one year ago
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    this is so frustrating, please help me understand this.

  33. mathcalculus
    • one year ago
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    @asnaseer

  34. phi
    • one year ago
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    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
    • one year ago
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    at the point (2,1)

  36. mathcalculus
    • one year ago
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    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
    • one year ago
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    it frustrated me to see this and not know how they even possibly do it.

  38. phi
    • one year ago
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    are you asking how to get from \[ \frac{-4x}{-2y} = \frac{2x}{y} \] ? -4/-2 is +2

  39. mathcalculus
    • one year ago
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    yes but with diferet variables?? 0.o

  40. mathcalculus
    • one year ago
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    different*

  41. phi
    • one year ago
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    what is the question ?

  42. mathcalculus
    • one year ago
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    oh the slope is -1/4! =]

  43. phi
    • one year ago
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    yes. also, the tangent line goes through point (2,1) so you can find the intercept.

  44. mathcalculus
    • one year ago
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    cool, how do you know this?

  45. mathcalculus
    • one year ago
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    okie doke

  46. mathcalculus
    • one year ago
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    thanks @phi you helped me so much! ^_^ so thankful

  47. phi
    • one year ago
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    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

  48. mathcalculus
    • one year ago
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    y=-1/4x+3/2

  49. phi
    • one year ago
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    yes

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