find \\( \\frac { d y } { d x } \\) for \\( y = \\sqrt { u } \\) and \\( u = x ^ { 2 } + 1 \\). state your…

Explanation:

Step1: Find (\frac{dy}{du})

Given (y = \sqrt{u}=u^{\frac{1}{2}}). Using the power rule (\frac{d}{du}(u^n)=nu^{n - 1}), we have (\frac{dy}{du}=\frac{1}{2}u^{\frac{1}{2}-1}=\frac{1}{2}u^{-\frac{1}{2}}=\frac{1}{2\sqrt{u}}).

Step2: Find (\frac{du}{dx})

Given (u=x^{2}+1). Using the power rule (\frac{d}{dx}(x^n)=nx^{n - 1}) and (\frac{d}{dx}(c)=0) (where (c) is a constant), we get (\frac{du}{dx}=2x+0 = 2x).

Step3: Use the chain - rule (\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx})

Substitute (\frac{dy}{du}=\frac{1}{2\sqrt{u}}) and (\frac{du}{dx}=2x) into the chain - rule formula. (\frac{dy}{dx}=\frac{1}{2\sqrt{u}}\cdot2x). Since (u = x^{2}+1), we substitute (u) with (x^{2}+1). (\frac{dy}{dx}=\frac{x}{\sqrt{x^{2}+1}}).

Answer:

(\frac{x}{\sqrt{x^{2}+1}})