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

1. anonymous

$y'=-9x ^{2} \cos(-x ^{3})$

2. anonymous

3. anonymous

it is the derivative of 3sin(u) times the derivative of u where $u=-x ^{3}$

4. 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

5. anonymous

two answers. which one is right

6. 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)

7. 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!!

8. anonymous

ok i get ur wrk so whats the final answer

9. 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

10. anonymous

the only thing that wouldn't be understandable is the - sign. that's a negative and not a minus

11. amistre64

1-x^6

12. amistre64

1-(x^6) is that more clear?

13. anonymous

yes hey thx. would u like another prob

14. amistre64

lol.... sure, but I dont know how much help I would be :)

15. 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?"

16. 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?

17. 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

18. amistre64

dc/dt = [a(da/dt) + b(db/dt)] / c

19. amistre64

dc/dt = derivative of (a^2 + b^2) with respect to "t"; divided by "c"

20. amistre64

make that ... divided by "2c"

21. amistre64

let me know if you think I am wrong :) and why I would be wrong

22. anonymous

i see what you are doin, maybe the answer i get is wrong. brb