# Sin/cos inverse - derivitives

1. Nov 7, 2005

### Yura

I'm studying for my exam which is in four days and I've searched through my text book (im using an old book because thats the one the school uses) and I cant seem find how to do the derivitives for the sine/cosine inverse functions:

sin^(-1)x

cos^(-1)x

Thanks.

2. Nov 7, 2005

### Kamataat

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

- Kamataat

3. Nov 7, 2005

### Kristi

sin¯¹x=1/√1-x² and cos¯¹x=-1/√1-x²

Hope this is what you are looking for.

4. Nov 7, 2005

### 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

Last edited: Nov 7, 2005
5. Nov 7, 2005

### Yura

Ah, I get it now. Thanks everyone!

6. Nov 7, 2005

### 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.

7. Nov 7, 2005

### 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.

8. Nov 13, 2005

### 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.