# Math Help - differential

1. ## differential

y=sinx+cosx is a function defined in the interval [0,pi/6], find derivative of f^-1(x)?

2. ## Re: differential

The derivative of $f^{-1}(x)$ is $\frac{1}{f'(x)}$.

- Hollywood

3. ## Re: differential

Yes it is but I couldn't find the answer given by this method.

4. ## Re: differential

y^2=1+sin2x y=sqrt(1+sin2x) and if we differentiate y, and write it

1/dy/dx, we find 1/cosx/sqrt(1+sin2x).

5. ## Re: differential

You have $y=\sin{x}+\cos{x}$ so $\frac{dy}{dx}$ should be $\cos{x}-\sin{x}$. I don't see how you get $\frac{1}{\sqrt{2-x^2}}$. Are you sure the problem and answer are both copied correctly?

I follow you on the second line of your post. On the third line you should have $\cos{2x}$ instead of $\cos{x}$ giving $\frac{\sqrt{1+\sin{2x}}}{\cos{2x}}$.

- Hollywood

6. ## Re: differential

[QUOTE=hollywood;787266]You have $y=\sin{x}+\cos{x}$ so $\frac{dy}{dx}$ should be $\cos{x}-\sin{x}$. I don't see how you get $\frac{1}{\sqrt{2-x^2}}$. Are you sure the problem and answer are both copied correctly?

I follow you on the second line of your post. On the third line you should have $\cos{2x}$ instead of $\cos{x}$ giving $\frac{\sqrt{1+\sin{2x}}}{\cos{2x}}$.
y^2=1+sin2x, f^-1(x)=1/2.arcsin(x^2-1),x/(2x^2-x^4)

7. ## Re: differential

Oh, sure enough. It should be $\frac{x}{\sqrt{2x^2-x^4}}$, and then you can cancel x giving $\frac{1}{\sqrt{2-x^2}}$.

In my solution I forgot to swap y and x, so I should have had $\cos{y}-\sin{y}$, which gives the same thing if you substitute $y=\frac{1}{2}\sin^{-1}(x^2-1)$ (after a lot of algebra). Your solution is more direct. Well done.

- Hollywood