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anonymous
 5 years ago
find derivative of y w/ respect to x. y=3 sin^1 x^3
anonymous
 5 years ago
find derivative of y w/ respect to x. y=3 sin^1 x^3

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[y'=9x ^{2} \cos(x ^{3})\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i'm interested in knowing how you got to that answer please

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it is the derivative of 3sin(u) times the derivative of u where \[u=x ^{3}\]

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0y = 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(1x^6)) * 3x^2 = (6x^2) / (sqrt(1x^6)) if I recall correctly

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0two answers. which one is right

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0the 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(1sin^2(sin1(x))) the derivative of sin^1(x) = 1/sqrt(1x^2)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0the derivative of sin^1(u) = 1/sqrt(1u^2); where u=x^3 1/sqrt(1(x^3^2)) = 1/sqrt(1x^6) so, 3 * 3x^2 * 1/sqrt(1x^6) = (9x^2)/(sqrt(1x^6)). I forgot how to multiply 3 times 3 :) its 9!!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok i get ur wrk so whats the final answer

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0the final answer that I get, as long as I understood your original equation is: (9x^2) / sqrt(1x^6) the sqrt means square root

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the only thing that wouldn't be understandable is the  sign. that's a negative and not a minus

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.01(x^6) is that more clear?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes hey thx. would u like another prob

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0lol.... sure, but I dont know how much help I would be :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you'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?"

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0Lets 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?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so 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

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0dc/dt = [a(da/dt) + b(db/dt)] / c

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0dc/dt = derivative of (a^2 + b^2) with respect to "t"; divided by "c"

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0make that ... divided by "2c"

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0let me know if you think I am wrong :) and why I would be wrong

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i see what you are doin, maybe the answer i get is wrong. brb
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