# using chain rule with circular functions

I have the following equation: $$\sin^{-1}\left(\frac{x-1}{2}\right)$$

Is it ok to solve like this:

$y=\sin^{-1}(z)$, $z=\frac{x-1}{2}$

$\frac{dy}{dz} = \frac{1}{\sqrt{1-z^2}}$ and $\frac{dz}{dx}=\frac{2-(x-1)}{4}$

$\frac{dy}{dx}=\frac{dy}{dz}\frac{dz}{dx}$.

Here I do not know what to do more. Please tell me if I am doing the things above correct, and how to finalize the answer. Thanks.

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The derivate of( x-1)/2 is 1/2/.You can apply the chain rule for all x such that ( x-1)/2 is biger than -1 and lower than 1, since there arcsinx is derivable.This is when x is bigger than -1 and lower than 3.At those points the dreivates laterals does not exist. – alpha.Debi Apr 12 '12 at 14:18
Your $dz\over dx$ is not correct, ${dz\over dx}={1\over2}$. Then just write the answer: ${dy\over dx}={dy\over dz}{dz\over dx}={1\over \sqrt {1-z^2}}\cdot {1\over2} ={1\over \sqrt {1-({x-1\over 2})^2}}\cdot {1\over2}$. – David Mitra Apr 12 '12 at 14:18
Can you please write how it will be 1/2 ? – Sean87 Apr 12 '12 at 14:19
The derivate of the denominator(2) is zero. – alpha.Debi Apr 12 '12 at 14:24
Your question is in some ways not clearly expressed. You say you have an equation, but what you write is not an equation. Then you speak of "solving" it, without saying what you're actually trying to do with it. Later, it turns out that what you want to do is to find its derivative with respect to $x$. But it could have been that what you wanted to do was something else, e.g. simplify it in some specified way. So you could have said "I have this function: $\cdots\cdots\cdots$. Is it OK to find its derivative like this: $\cdots\cdots\cdots$? – Michael Hardy Apr 12 '12 at 15:35

I'm assuming you want to find the derivative of $\sin^{-1}{x-1\over2}$.
As you did, you'll apply the chain rule. But, your $dz\over dx$ is not correct. If you use the quotient rule to find this, note the derivative of the denominator is zero; so $${dz\over dx}={d\over dx}{x-1\over 2}= {1\cdot2-(x-1)\cdot0\over 2^2}={1\over 2 }.$$ But, note, there is no need to use the quotient rule when the denominator is a constant: $${dz\over dx}={d\over dx}{x-1\over 2}= {1\over2}{d\over dx} (x-1)= {1\over2} (1-0)={1\over2} .$$
You calculated ${dy\over dz}$ correctly. Now, just write the answer, and don't forget to write everything in terms of $x$: $${dy\over dx}={dy\over dz}{dz\over dx}={1\over \sqrt {1-z^2}}\cdot {1\over2} ={1\over 2\sqrt {1-({x-1\over 2})^2}} .$$