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anonymous

  • 5 years ago

find derivative of y w/ respect to x. y=3 sin^-1 x^3

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  1. anonymous
    • 5 years ago
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    \[y'=-9x ^{2} \cos(-x ^{3})\]

  2. anonymous
    • 5 years ago
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    i'm interested in knowing how you got to that answer please

  3. anonymous
    • 5 years ago
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    it is the derivative of 3sin(u) times the derivative of u where \[u=-x ^{3}\]

  4. amistre64
    • 5 years ago
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    y = 3sin^-1(x^3) (d/dx)(y) = (d/dx)(3sin^-1(x^3)) (dy/dx) = 3 (dx/dx) (sin^-1(x^3)) = 3 (1/sqrt(1-x^6)) * 3x^2 = (6x^2) / (sqrt(1-x^6)) if I recall correctly

  5. anonymous
    • 5 years ago
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    two answers. which one is right

  6. amistre64
    • 5 years ago
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    the derivative of y=sin^1(x): sin(y) = x (d/dx)(sin(y)) = (d/dx)(x) (dy/dx)(cos(y)) = (dx/dx) y' = 1/cos(y) y' = 1/(cos(sin^-1(x))) y' = 1/sqrt(1-sin^2(sin-1(x))) the derivative of sin^-1(x) = 1/sqrt(1-x^2)

  7. amistre64
    • 5 years ago
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    the derivative of sin^-1(u) = 1/sqrt(1-u^2); where u=x^3 1/sqrt(1-(x^3^2)) = 1/sqrt(1-x^6) so, 3 * 3x^2 * 1/sqrt(1-x^6) = (9x^2)/(sqrt(1-x^6)). I forgot how to multiply 3 times 3 :) its 9!!

  8. anonymous
    • 5 years ago
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    ok i get ur wrk so whats the final answer

  9. amistre64
    • 5 years ago
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    the final answer that I get, as long as I understood your original equation is: (9x^2) / sqrt(1-x^6) --the sqrt means square root

  10. anonymous
    • 5 years ago
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    the only thing that wouldn't be understandable is the - sign. that's a negative and not a minus

  11. amistre64
    • 5 years ago
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    1-x^6

  12. amistre64
    • 5 years ago
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    1-(x^6) is that more clear?

  13. anonymous
    • 5 years ago
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    yes hey thx. would u like another prob

  14. amistre64
    • 5 years ago
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    lol.... sure, but I dont know how much help I would be :)

  15. anonymous
    • 5 years ago
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    you've been enough help so far. k here it is. "if a and b are lengths of the legs of a right triangle and c is the length of the hypotenuse, then c^2 = a^2 + b^2. How is dc/dt related to da/dt and db/dt?"

  16. amistre64
    • 5 years ago
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    Lets solve the equation implicitly with respect to time. c^2 = a^2 + b^2 (d/dt) (c^2) = (d/dt)(a^2) + (d/dt)(b^2) (dc/dt)(2c) = (da/dt)(2a) + (db/dt)(2b) To clean it up alittle, lets use c' = (dc/dt) and similar notation for the others. c'2c = a'2a + b'2b divide everything by 2 to simply: c'(c) = a'(a) + b'(b) Now, as long as you know your abc's a'b'c', you solve algebraiclly for the unknowns. Does that make sense?

  17. anonymous
    • 5 years ago
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    so the relationship is pretty much the same. the result would be dc/dt is the derivative of (da/dt)^2 + (db/dt)^2. correct me if i'm wrong

  18. amistre64
    • 5 years ago
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    dc/dt = [a(da/dt) + b(db/dt)] / c

  19. amistre64
    • 5 years ago
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    dc/dt = derivative of (a^2 + b^2) with respect to "t"; divided by "c"

  20. amistre64
    • 5 years ago
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    make that ... divided by "2c"

  21. amistre64
    • 5 years ago
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    let me know if you think I am wrong :) and why I would be wrong

  22. anonymous
    • 5 years ago
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    i see what you are doin, maybe the answer i get is wrong. brb

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