anonymous
  • anonymous
find derivative of y w/ respect to x. y=3 sin^-1 x^3
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
\[y'=-9x ^{2} \cos(-x ^{3})\]
anonymous
  • anonymous
i'm interested in knowing how you got to that answer please
anonymous
  • anonymous
it is the derivative of 3sin(u) times the derivative of u where \[u=-x ^{3}\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

amistre64
  • amistre64
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
anonymous
  • anonymous
two answers. which one is right
amistre64
  • amistre64
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)
amistre64
  • amistre64
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!!
anonymous
  • anonymous
ok i get ur wrk so whats the final answer
amistre64
  • amistre64
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
anonymous
  • anonymous
the only thing that wouldn't be understandable is the - sign. that's a negative and not a minus
amistre64
  • amistre64
1-x^6
amistre64
  • amistre64
1-(x^6) is that more clear?
anonymous
  • anonymous
yes hey thx. would u like another prob
amistre64
  • amistre64
lol.... sure, but I dont know how much help I would be :)
anonymous
  • anonymous
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?"
amistre64
  • amistre64
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?
anonymous
  • anonymous
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
amistre64
  • amistre64
dc/dt = [a(da/dt) + b(db/dt)] / c
amistre64
  • amistre64
dc/dt = derivative of (a^2 + b^2) with respect to "t"; divided by "c"
amistre64
  • amistre64
make that ... divided by "2c"
amistre64
  • amistre64
let me know if you think I am wrong :) and why I would be wrong
anonymous
  • anonymous
i see what you are doin, maybe the answer i get is wrong. brb

Looking for something else?

Not the answer you are looking for? Search for more explanations.