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.