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Lammy

  • 4 years ago

find the equation of the tangent to the curve y=X^(x^2) - X^(Pi) where x=1 find the equation of the tangent to the curve y=X^(x^2) - X^(Pi) where x=1 @Mathematics

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  1. Annand
    • 4 years ago
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    Take log of y and then differentiate... (dy/dx) at x=1 will give the slope of the tangent...

  2. Lammy
    • 4 years ago
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    when you say dy/dx at x=1 do you mean to plug in 1 for all x after i get the derivative of the equation?

  3. Annand
    • 4 years ago
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    exactly

  4. Lammy
    • 4 years ago
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    is the slop -10.6063

  5. Annand
    • 4 years ago
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    you can also find y at x=1... Thus you know the slope and one point in the line which will help you to find the equation...

  6. Annand
    • 4 years ago
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    what was your dy/dx

  7. Lammy
    • 4 years ago
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    http://www.wolframalpha.com/input/?i=derivative+x%5E%28x%5E2%29-x%5E%28%CF%80%29

  8. Annand
    • 4 years ago
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    is it not (1-pi)?

  9. Lammy
    • 4 years ago
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    yes yes thats the slope right? then how do i find the y to plug in for point slope form

  10. Annand
    • 4 years ago
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    you have the equation of the curve... put x=1 in it... that'll give you the point required...

  11. Annand
    • 4 years ago
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    that is y=X^(x^2) - X^(Pi) at x=1

  12. Lammy
    • 4 years ago
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    = 0

  13. Annand
    • 4 years ago
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    yep

  14. Annand
    • 4 years ago
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    so you've got a point and the slope... go on...

  15. Lammy
    • 4 years ago
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    can i leave it at y-0=1-pi(x-1)

  16. Annand
    • 4 years ago
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    y-0=(1-pi)(x-1) don't forget the brackets...

  17. Lammy
    • 4 years ago
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    thats it?

  18. Annand
    • 4 years ago
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    you may write it in the form y=mx+c if you like...

  19. Lammy
    • 4 years ago
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    y=-piX +X+pi-1

  20. Annand
    • 4 years ago
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    y=(1-pi)x-(1-pi)

  21. Lammy
    • 4 years ago
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    thank you soo much!

  22. Annand
    • 4 years ago
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    you're welcome...

  23. Annand
    • 4 years ago
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    A piece of advice: you should've calculated the derivative by hand....

  24. Lammy
    • 4 years ago
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    yea thats the problem i dont know how to do that one

  25. Annand
    • 4 years ago
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    do you want me to explain?

  26. Lammy
    • 4 years ago
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    yes please

  27. Annand
    • 4 years ago
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    In y=X^(x^2) - X^(Pi) the problem is with the first term... I believe that you know how to differentiate x^(pi)...

  28. Lammy
    • 4 years ago
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    yea

  29. Annand
    • 4 years ago
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    let y=p-q where p=X^(x^2) and q=X^(Pi)

  30. Annand
    • 4 years ago
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    taking log of p log(p)=x^2*log(x) then differentiate (1/p)dp/dx=x^2/x+2xlog(x) dp/dx=p[x+2xlog(x)] dp/dx=x^(x^2)*[x+2xlog(x)]

  31. Annand
    • 4 years ago
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    y=dp/dx-dq/dx

  32. Annand
    • 4 years ago
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    got it?

  33. Lammy
    • 4 years ago
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    (1/p)dp/dx=x^2/x+2xlog(x)

  34. Lammy
    • 4 years ago
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    so derivative of log(x) = 2xlog(x)

  35. Annand
    • 4 years ago
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    nope... derivative of log(x) is 1/x

  36. Annand
    • 4 years ago
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    p is the product of x^2 and log(x) so you've to use the product rule.

  37. Annand
    • 4 years ago
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    \[d(x^2*\log(x))/dx=x^2*d(\log(x))/dx + \log(x)*d(x^2)/dx\]

  38. Lammy
    • 4 years ago
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    where is the 1/p from? (1/p)dp/dx=x^2/x+2xlog(x)

  39. Lammy
    • 4 years ago
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    where did the x^2 go? dp/dx=p[x+2xlog(x)]

  40. Annand
    • 4 years ago
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    x^2/x=x

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