Question

f(x)=sin^(−1)(x^2)−(1/sin^2x)+2sin(2x)

Answer 1

Need to find the derivative of that at -pi/4

Answer 2

\[f(x)=\sin^{-1}x^2-(1/\sin^2x)+2\sin(2x)\] That might be clearer

Answer 3

Well, first of all find the derivative, then substitute -Pi/4

\[f'(x)=\frac{ 2x }{ \sqrt{1-x^4} }-2cscx *(-cscx*cotx)+2\cos(2x)*2\]
\[f'(x)=\frac{ 2x }{ \sqrt{1-x^4} }+2\csc(x)^2*\cot(x)+4\cos(2x)\]

Answer 4

Now you only have to substitute -Pi/4, but i think the hardest part was the derivative. Do you have doubts with the substitution?

Answer 5

How did you get the derivative for sin^-1(x^2)? The way I did it was to first rewrite it as (sinx^2)^-1, took the derivative making it (-1)(sin(x^2))^-2 times cos(x^2) times 2x (chain rule).

Answer 6

@Kikazo

Answer 7

Oh well, then I might have misunderstood. Or maybe you have. As far as i know, writing \[\sin ^{-1}(x)\] refers to the inverse sin function: arcsin(x). Because \[(\sin(x))^{-1}= \frac{ 1 }{ sinx }=cscx\] so you would write cscx instead of sin^(-1)x

Answer 8

I see... I think you are probably right. In which case what is the general derivative of sin^-1(x)?

Answer 9

\[\frac{ d \arcsin(u) }{ dx }=\frac{ u' }{ \sqrt{1-u^2} }\]

Answer 10

You can verify that in any calculus book n.n

Answer 11

Yeah I got that same derivative from wolfram. So then it would be \[2x/(\sqrt{1-x^2})\]

Answer 12

nope, it would be \[\frac{ 2x }{ \sqrt{1-x^4} }\] , for you have to square the argument

Answer 13

Awesome! So the derivative for 1/sin^2(x) is [-2sinxcosx]/[sin^4(x)]?

Answer 14

No, for that part... i decided to rewrite it. Since cscx=1/sinx, 1/sin^2x=csc^2(x). Then use the formula, 2*cscx*(-cscx*cotx)

Answer 15

Or you can leave it as it is, and find the derivattive using the u/v formula

Answer 16

\[-4+(-\pi/2)/\sqrt{1-(\pi^4/256)}\]Is what I got as the derivative at -pi/4

Answer 17

Yup, I tried the problem again using the formula you gave (also one that I came up with on my own) and got the same answer twice

Answer 18

i'm not quite sure about that -4, let me check...

Answer 19

yeah, I think it's ok :D

Answer 20

thanks a ton