Thx for the help.
$\displaystyle y'=\frac{1}{2}(1-sin\;x)^{-\frac{1}{2}}(-cos\;x)$ at zero...u just plug in zero now into the derivative.
You should get $\displaystyle -\frac{1}{2}$
As for the domain, you know that $\displaystyle sin\;x$ will always have a value between 0 to 1 so 1-(minus)anything from zero to 1 will always get you a real number, therefore the doman is xER(all real number)
$\displaystyle y = \sqrt{1 - sin(x)}$
First the domain. The domain of sin(x) is all real numbers, so the sin(x) puts no restrictions on the value of x. The next thing to look at is the square root; the argument of a square root can never be negative. But 1 - sin(x) is never negative. So the domain of y is all real numbers.
As far as the tangent line:
$\displaystyle y^{\prime} = \frac{1}{2} \cdot \frac{1}{\sqrt{1 - sin(x)}} \cdot -cos(x)$
I'll leave the task of simplifying this and letting x = 0 to you.
-Dan
Okay. Well, we have the same arguments as I used before, but now we need to worry about the denominator being 0. So we have the requirement that
$\displaystyle \sqrt{1 - sin(x)} \neq 0$
$\displaystyle 1 - sin(x) \neq 0$
$\displaystyle sin(x) \neq 1$
So what values of x are excluded from the domain?
-Dan