Question

need some to show me step by step on how to do the attachment problem

Answer 1

i'll just check out derivative of arcsin x/3 - to make sure

Answer 2

there should be no negative in front of, 2arcsin(x/3)

Answer 3

2 arcsin (x/3) right?

Answer 4

yes

Answer 5

derivative of arcsin x/3 = 1 / sqrt (3^2 - x^2) = 1 / sqrt(9 - x^2)
sorry - my memory failed me there

so the correct answer is 2 / sqrt (9 - x^2)

option 4

Answer 6

forget chain rule thing

Answer 7

how do did you get 3^(2) inside the square root?

Answer 8

its a standard derivative

derivative of arcsin x/a = 1 / sqrt (a^2 - x^2)

Answer 9

ok is that always the case for the other trig function like arctan x/a, arccos x/a, arccot x/a, arccsc x/a, arcsec x/a? Also what if the problem was arcsin(2x) what would that answer be?

Answer 10

because in the question you just did there was fraction inside the trig function, but what if it was a whole number like 2X instead of x/3

Answer 11

ok - yes there are standard forms for arcos, arctan etc

Answer 12

because the think that confusing me that the derivative of arcsin is 1/sqrt(1-x^(2)), so in this problem you treat the 1 in the square root like a

Answer 13

also in this problem what happen to the x/3 once you take the derivative which should equal 1/3? The chain rule is when take the derivative of the outside first then the inside.

Answer 14

in the case of arcsin 2x you would use the chain rule
derivative of 2x is 2
derivative of arcsin u is 1 / sqrt(1 - u^2) u = 2x
so derivative is

2 * 1 / sqrt (1 - (2x)^2) = 2 / sqrt (1 - 4x^2)

Answer 15

for arcsin x the a is replaced by 1 so its 1 / sqrt ( 1^2 - x^2)

Answer 16

so are not suppose to take the derivative of the inside? because it seems like your just taking out a 3 from the inside.

Answer 17

Also when you said u=2x shouldn't you divide the 2 to the other side so the x can be by itself the substitution to work, u/2=x

Answer 18

no the 'inside' function is 2x and derivative of 2x is 2
let me illustrate the chain rule with a simpler problem

f(x) = sin 3x find f'(x)

let y = sin 3x
let u = 3x so du/dx = 3
y = sin u so dy/du = cos u

therefore dy/dx = dy/du * du/dx = 3 cos u

now replace u by 3x :-

dy/dx = 3 cos 3x

Answer 19

maybe u prefer to use the f' notation than dy/dx - i find the dy/dx notation easier

Answer 20

especially when it comes to thing like chain rule

Answer 21

yeah, but the only thing that is confusing me is the number that are inside the trig function like when you had x/3 we decide for the 3 on the bottom a, but we never took the derivative of it. for the problem that was 2x we took the derivative of it then used u-substitution and totally ignore the a in the square root

Answer 22

right - i see your problem - i could have solved the original problem using the chain rule - letting u = x/3 but it would have involved tedious algebra - its much easier to use the standard forms. For example you can differentiate sin x from first principles but its far easier to use standard form cos x.

Answer 23

i just forgot the standard form!

Answer 24

for sin 2x, why is a not equal 2 and why is the 2 on top?

Answer 25

you didn't use standard-form for arcsin2x

Answer 26

no - well theres not a standard form for every function - and it wasnt a difficult one to solve

Answer 27

i dont know who decides standard forms...

Answer 28

so how do you know when to use standard-forms

Answer 29

but it makes sense in the case of arcsin x/a etc because they are awkward to do

Answer 30

well commit them to memory - they are in all the textbooks

Answer 31

I guess, but thanks for helping and trying to explain thing out to me

Answer 32

no problem

Answer 33

calculus is not the easiest of subjects

Answer 34

you kidding, lol

Answer 35

lol