What are the derivatives of sine and cosine inverse functions?

[Yura]

I'm studying for my exam which is in four days and I've searched through my textbook (im using an old book because that's the one the school uses) and I can't seem find how to do the derivitives for the sine/cosine inverse functions:

sin^(-1)x

cos^(-1)x

Could anyone please tell me what these derive to please?
Thanks.

 

[Kamataat]

Use {y'}_x = \frac{1}{{x'}_y}. Take y=\arcsin x and x=\sin y.

- Kamataat

 

[Kristi]

sin¯¹x=1/√1-x² and cos¯¹x=-1/√1-x²

Hope this is what you are looking for.

 

[kreil]

I'll derive the sine one for you, try to use the process i use to find cosine.

g'(x)=\frac{1}{f'(g(x))}

f(x)=sin(x), g(x)=sin^-1(x)

and

y=sin(x) implies that x=sin(y) for the inverse


We know the derivative of sine is cosine, so the equation becomes..

\frac{1}{cos(sin^{-1}(x))}

but we know that

cos(y)=\sqrt{1-sin^2(y)}

also using the idea that cos(sin^{-1}(x))=cos(y)

\frac{1}{\sqrt{1-sin^2y}}

and from an equation above, x=sin(y), so the final product is:

\frac{d}{dx}(sin^{-1}(x))=\frac{1}{\sqrt{1-x^2}}

Josh

 

[Yura]

Ah, I get it now. Thanks everyone!

 

[dextercioby]

It's too complicated.

Differentiate in both sides wrt "x"

x=\sin\left(\arcsin x\right)

and

x=\cos\left(\arccos x\right)

using for the RHS the chain rule.

Daniel.

 

[Jameson]

I think everyone's is too complicated. Here's how I derive it.

(1)y=\sin^{-1}(x)

(2)\sin(y)=x

(3)\cos(y)\frac{dy}{dx}=1

(4)\frac{dy}{dx}=\frac{1}{\cos(y)}

Use a right trianlge to evaluate the cosine of y. So it's four written steps and however many you want to call using the triangle.

 

[Robokapp]

They are positive(for sin) and negative(for cos) 1/[(1-x^2)^(1/2)]

Unfortunatelly I had to memorize them...I have no clue why they're that, but they're 1 divided by square root of 1 - x^2 and negative that for cos.