Problem

Find $\frac{d y}{d x}$ for $y=\sqrt{u}$ and $u=x^{2}+1$.
\[
\frac{d y}{d x}=
\]

Answer

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Answer

Finally, we check the result. The derivative \(\frac{x}{\sqrt{x^{2}+1}}\) is defined for all real numbers \(x\), and it is in the simplest form. So, the result meets the requirements of the problem.

Steps

Step 1 :Given the function \(y=\sqrt{u}\) and \(u=x^{2}+1\), we want to find \(\frac{d y}{d x}\).

Step 2 :First, we can find \(\frac{d y}{d u}\) and \(\frac{d u}{d x}\).

Step 3 :For \(y=\sqrt{u}\), \(\frac{d y}{d u}=\frac{1}{2\sqrt{u}}\).

Step 4 :For \(u=x^{2}+1\), \(\frac{d u}{d x}=2x\).

Step 5 :Then, we can use the chain rule to find \(\frac{d y}{d x}\). The chain rule states that \(\frac{d y}{d x}=\frac{d y}{d u} \cdot \frac{d u}{d x}\).

Step 6 :Substituting \(\frac{d y}{d u}=\frac{1}{2\sqrt{u}}\) and \(\frac{d u}{d x}=2x\) into the chain rule, we get \(\frac{d y}{d x}=\frac{1}{2\sqrt{u}} \cdot 2x\).

Step 7 :Substitute \(u=x^{2}+1\) into the equation, we get \(\frac{d y}{d x}=\frac{1}{2\sqrt{x^{2}+1}} \cdot 2x\).

Step 8 :Simplify the equation, we get \(\frac{d y}{d x}=\frac{x}{\sqrt{x^{2}+1}}\).

Step 9 :So, the derivative of \(y=\sqrt{u}\) with respect to \(x\) where \(u=x^{2}+1\) is \(\frac{x}{\sqrt{x^{2}+1}}\).

Step 10 :Finally, we check the result. The derivative \(\frac{x}{\sqrt{x^{2}+1}}\) is defined for all real numbers \(x\), and it is in the simplest form. So, the result meets the requirements of the problem.

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