Question

Derivative. Is this correct?
f(x)=arcsin(x/cos(x))
(formatting and work in reply)

Answer 1

\[f(x)=\arcsin(\frac{ x }{ \cos(x) })\]
\[f'(x)=\frac{ 1 }{ \sqrt{1+(\frac{ x }{ \cos(x) }})^2 }(\frac{ \cos(x)-x(-\sin(x)) }{ \cos(x)^2 })\]
I didn't simplify much, so it's easier to see first if I'm doing it right.

Answer 2

whatever you are doing, i have a suggestion
as ugly as this sadistic problem is going to be, it is going to be easier to start with
\[\arcsin(x\sec(x))\]

Answer 3

only mistake i see is that you should have a minus sign inside the radical

Answer 4

\[f'(x)=\frac{ 1 }{ \sqrt{1-(\frac{ x }{ \cos(x) }})^2 }(\frac{ \cos(x)-x(-\sin(x)) }{ \cos(x)^2 })\]

Answer 5

it would make it a lot easier to write if you use \(x\sec(x)\)
the answer will be the same though
you are correct

Answer 6

ah yeah, that should be minus.
ok good. I double checked with wolfram, but it simplifies to the max and it looked so different, I wasn't sure if I was doing it right.
http://www.wolframalpha.com/input/?i=derivative+of+arcsin%28x%2Fcos%28x%29%29

Answer 7

i don't really think it is a matter of taking your answer (which is correct) and using a bunch of trig formulas
if you replace \(\frac{x}{\cos(x)}\) by \(x\sec(x)\) you can use the product rule rather than the quotient rule

Answer 8

the first part is the same, the second would be
\[\sec(x)+x\sec(x)\tan(x)\]

Answer 9

I see, because sec is 1/cos. So x/cos would be x sec (x).

Answer 10

then as wolfram did you can factor out the \(\sec(x)\) so that the numerator will be
\[\sec(x)(\tan(x)+1)\]

Answer 11

it is just that the product rule comes out a lot neater than the quotient rule, but again everything you wrote (except the plus sign ) is correct

Answer 12

I never would have thought that it would be easier dealing with sec than cos. I just worked it out with sec and see that yeah, it is a lot neater because of being able to use that product rule instead of the quotient rule. Thanks!