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

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- May 15th 2013, 05:58 AMkastamonudifferential
y=sinx+cosx is a function defined in the interval [0,pi/6], find derivative of f^-1(x)?

- May 15th 2013, 07:08 AMhollywoodRe: differential
The derivative of $\displaystyle f^{-1}(x)$ is $\displaystyle \frac{1}{f'(x)}$.

- Hollywood - May 15th 2013, 10:01 PMkastamonuRe: differential
Yes it is but I couldn't find the answer given by this method.

- May 15th 2013, 10:12 PMkastamonuRe: differential
I couldn't find the answer. Answer is 1/sqrt(2-x^2)

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). - May 16th 2013, 06:52 AMhollywoodRe: differential
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 - May 16th 2013, 07:34 AMkastamonuRe: differential
[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) - May 16th 2013, 08:17 AMhollywoodRe: differential
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