y=sinx+cosx is a function defined in the interval [0,pi/6], find derivative of f^-1(x)?
You have $\displaystyle y=\sin{x}+\cos{x}$ so $\displaystyle \frac{dy}{dx}$ should be $\displaystyle \cos{x}-\sin{x}$. I don't see how you get $\displaystyle \frac{1}{\sqrt{2-x^2}}$. 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 $\displaystyle \cos{2x}$ instead of $\displaystyle \cos{x}$ giving $\displaystyle \frac{\sqrt{1+\sin{2x}}}{\cos{2x}}$.
- Hollywood
[QUOTE=hollywood;787266]You have $\displaystyle y=\sin{x}+\cos{x}$ so $\displaystyle \frac{dy}{dx}$ should be $\displaystyle \cos{x}-\sin{x}$. I don't see how you get $\displaystyle \frac{1}{\sqrt{2-x^2}}$. 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 $\displaystyle \cos{2x}$ instead of $\displaystyle \cos{x}$ giving $\displaystyle \frac{\sqrt{1+\sin{2x}}}{\cos{2x}}$.
y^2=1+sin2x, f^-1(x)=1/2.arcsin(x^2-1),x/(2x^2-x^4)
Oh, sure enough. It should be $\displaystyle \frac{x}{\sqrt{2x^2-x^4}}$, and then you can cancel x giving $\displaystyle \frac{1}{\sqrt{2-x^2}}$.
In my solution I forgot to swap y and x, so I should have had $\displaystyle \cos{y}-\sin{y}$, which gives the same thing if you substitute $\displaystyle y=\frac{1}{2}\sin^{-1}(x^2-1)$ (after a lot of algebra). Your solution is more direct. Well done.
- Hollywood