9. find the derivative of the function ( y=\tan ^{-1} sqrt{x - 1} ). simplify the result. leave the result…

Explanation:

Step1: Apply the chain rule

Let (u = \sqrt{x - 1}), then (y=\tan^{- 1}(u)). The derivative of (\tan^{-1}(u)) with respect to (u) is (\frac{1}{1 + u^{2}}), and the derivative of (u=\sqrt{x - 1}=(x - 1)^{\frac{1}{2}}) with respect to (x) is (\frac{1}{2}(x - 1)^{-\frac{1}{2}}) by the power rule ((x^{n})^\prime=nx^{n - 1}). By the chain rule (\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}), we have (\frac{dy}{du}=\frac{1}{1 + u^{2}}) and (\frac{du}{dx}=\frac{1}{2\sqrt{x - 1}}).

Step2: Substitute (u) back

Since (u = \sqrt{x - 1}), then (1+u^{2}=1+(x - 1)=x). So (\frac{dy}{dx}=\frac{1}{1 + (\sqrt{x - 1})^{2}}\cdot\frac{1}{2\sqrt{x - 1}}).

Step3: Simplify the expression

Substitute (1+(\sqrt{x - 1})^{2}=x) into the above - expression, we get (\frac{dy}{dx}=\frac{1}{x}\cdot\frac{1}{2\sqrt{x - 1}}).

Answer:

(\frac{1}{2x\sqrt{x - 1}})