Question

please help ... .what is arccos(2/3x) ???

Answer 1

It is the measure of the angle whose cosine is 2/3x

Answer 2

You can't take it to any simpler evaluation because you still have x as a variable.

Answer 3

oooh okieee.

Answer 4

If you have a problem surrounding that expression, we look at the whole problem and see how it is used in its context.

Answer 5

the question is find the derivative of arcsin (3x/square root of 9x^2+4
so i drew a triangle to get cos^-1 which is 2/3x

Answer 6

and im not sure if this is the final answer or not.

Answer 7

\[\sin^{-1} \frac{ 3x }{ \sqrt{9x ^{2}+4} }\]Is that what you are working with? The derivative of that?

Answer 8

yup. so it said to make a right triangle to make it simplier to solve, so i did and i got cos to equal 2/3x.

Answer 9

Did your triangle look like this?

Answer 10

yayaya so i used cos to make the problem easier.

Answer 11

2 wasnt given, i found it out

Answer 12

Yes, I got it in also. What I did, was claculate the derivative without the triangle and got an answer, which is maybe what you are ultimately after anyway. I don't think I've used your technique is why I went the brute-force method, but I got an answer.

Answer 13

\[y = \tan^{-1} \frac{ 3x }{ 2 }\]I went and got the derivative both ways. First I did it with the original equation, what I called the "brute-force" method. I went back and with the triangle we were both using I used this new equation which is much much easier. Both ways gave the same derivative of \[\frac{ 6 }{ 9x ^{2}+4 }\]I believe that this is the substitution you were looking for. You still have to get (dy/du)(du/dx) with u=3x/2 but it is so much easier to work with arctan in this particular problem. If you have any more questions, don't hesitate to contact me. Sorry, but I couldn't get back to you earlier.