Question

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

Answer 1

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

Answer 2

i'm interested in knowing how you got to that answer please

Answer 3

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

Answer 4

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

Answer 5

two answers. which one is right

Answer 6

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)

Answer 7

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!!

Answer 8

ok i get ur wrk so whats the final answer

Answer 9

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

Answer 10

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

Answer 11

1-x^6

Answer 12

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

Answer 13

yes
hey thx. would u like another prob

Answer 14

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

Answer 15

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?"

Answer 16

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?

Answer 17

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

Answer 18

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

Answer 19

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

Answer 20

make that ... divided by "2c"

Answer 21

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

Answer 22

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