Thread: Quick question regarding tangent lines

1. Quick question

Thx for the help.

2. Originally Posted by Cursed
How do you find the line tangent to the function $y=sqrt(1-sinx)$ at x=0? Also, how do you find the domain of that function?
$y'=\frac{1}{2}(1-sin\;x)^{-\frac{1}{2}}(-cos\;x)$ at zero...u just plug in zero now into the derivative.

You should get $-\frac{1}{2}$

As for the domain, you know that $sin\;x$ 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)

3. Originally Posted by Cursed
How do you find the line tangent to the function $y=sqrt(1-sinx)$ at x=0? Also, how do you find the domain of that function?
$y = \sqrt{1 - sin(x)}$

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:
$y^{\prime} = \frac{1}{2} \cdot \frac{1}{\sqrt{1 - sin(x)}} \cdot -cos(x)$

I'll leave the task of simplifying this and letting x = 0 to you.

-Dan

4. Thanks.

This was my mistake, but I meant to ask what the domain was for the derivative. Could you help me with that? Thanks in advance.

5. Originally Posted by Cursed
Thanks.

This was my mistake, but I meant to ask what the domain was for the derivative. Could you help me with that? Thanks in advance.
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
$\sqrt{1 - sin(x)} \neq 0$

$1 - sin(x) \neq 0$

$sin(x) \neq 1$

So what values of x are excluded from the domain?

-Dan