Question

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Answer 1

Are you allowed to use the formula for derivative of arcsin(x)?

We can do it the long way if that's what you need.
Figure I'd ask first though.

Answer 2

We haven't learned to used derivatives with it yet so I guess we can go the short way? Lol. I got +/-csc(sqrt(x)) but it isn't one of the review choices.

Answer 3

\[\Large\bf\sf f(x)\quad=\quad \arcsin^2\left(\frac{1}{x}\right)\]\[\Large\bf\sf f'(x)\quad=\quad 2\arcsin\left(\frac{1}{x}\right)\cdot \color{royalblue}{\left[\arcsin\left(\frac{1}{x}\right)\right]'}\]
We start with power rule, then the chain rule will need to be applied.
That's what the blue part is all about.

Answer 4

That step make some sense? :U

Answer 5

Yes.

Answer 6

\[\large\bf\sf f'(x)\quad=\quad 2\arcsin\left(\frac{1}{x}\right)\cdot \color{orangered}{\frac{1}{\sqrt{1-\left(\dfrac{1}{x}\right)^2}}}\cdot \color{royalblue}{\left(\frac{1}{x}\right)'}\]Then arcsine gives us something like that, yes?
With another chain?

Answer 7

Then all we have left to do is to multiply it across right?

Answer 8

After taking the derivative of the second chain, yes.\[\large\bf\sf f'(x)\quad=\quad 2\arcsin\left(\frac{1}{x}\right)\cdot \color{orangered}{\frac{1}{\sqrt{1-\left(\dfrac{1}{x}\right)^2}}}\cdot \color{orangered}{\left(-\frac{1}{x^2}\right)}\]

Answer 9

(-2arcsin1/x) / x^2sqrt(1-1/x)^2

Answer 10

correct?

Answer 11

Careful careful, only the (1/x) is being squared in the denominator.

-2arcsin(1/x) / ( x^2sqrt(1-(1/x)^2) )

Answer 12

You can probably simplify this down a bit.
But if you're just trying to enter it online or something, I guess you don't need to worry about that.

Answer 13

Oh okay. Can you confirm another question with me? I'll make a new post if you want me to. I think it's pretty easy and I'm just overthinking it. lol.

Answer 14

sure :x