Question

find \\( \frac { d y } { d x } \\) for \\( y = \sqrt { u } \\) and \\( u = x ^ { 2 } + 1 \\). state your answer in terms of \\( x \\) only. \\( \frac { d y } { d x } = \\)

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:

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