Q. \( \frac{d}{dx}\left(x^{\frac{1}{2}}\right) \)

Q: What is the derivative of x^{1/2}?

Use the power rule: \dfrac{d}{dx}x^{n}=nx^{n-1}. Here n=\tfrac{1}{2}, so \dfrac{d}{dx}x^{1/2}=\tfrac{1}{2}x^{-1/2}=\dfrac{1}{2\sqrt{x}}, for x>0.

Q: How do you rewrite \sqrt{x} to differentiate it?

Rewrite x^{1/2}. Then apply the power rule: \dfrac{d}{dx}\sqrt{x}=\dfrac{d}{dx}x^{1/2}=\tfrac{1}{2}x^{-1/2}.

Q: What is the derivative of \sqrt{x} using limits/definition?

Start with f(x)=x^{1/2}. Compute f'(x)=\lim_{h\to 0}\dfrac{(x+h)^{1/2}-x^{1/2}}{h}. Rationalize, simplify, and get f'(x)=\tfrac{1}{2}x^{-1/2}.

Q: What is the domain of the derivative for x^{1/2}?

Since \sqrt{x} is real only for x\ge 0, the derivative formula \tfrac{1}{2}x^{-1/2} is valid for x>0. At x=0, it is undefined/infinite.

Q: What is the derivative of x^{1/2} for negative x (real-valued)?

For real calculus, \sqrt{x} is not defined when x<0. So the derivative \tfrac{1}{2}x^{-1/2} is not real-valued there.

Q: How to differentiate x^{1/2} using the chain rule?

If written as \sqrt{u} with u=x, then \dfrac{d}{dx}\sqrt{u}=\dfrac{1}{2\sqrt{u}}\cdot \dfrac{du}{dx}. With u=x, this gives \tfrac{1}{2\sqrt{x}}.

Q: Is \dfrac{d}{dx}x^{1/2}=\dfrac{1}{2\sqrt{x}} always correct?

It is correct for x>0. More generally, \tfrac{1}{2}x^{-1/2} matches the same derivative wherever x^{1/2} is differentiable in the chosen real domain.

Answer

Let \(f(x)=x^{1/2}\). Using the power rule \(\frac{d}{dx}x^n=nx^{n-1}\) with \(n=\tfrac{1}{2}\):

\[
\frac{d}{dx}x^{1/2}=\frac{1}{2}x^{-1/2}=\frac{1}{2\sqrt{x}}
\]

Detailed Explanation

We want to find the derivative of the function \(f(x)=x^{1/2}\).

Step 1: Identify the function form

\(f(x)=x^{1/2}\) is a power function of the form \(x^{n}\), where \(n=\tfrac{1}{2}\).

Step 2: Use the Power Rule

The Power Rule for derivatives says:

\[
\frac{d}{dx}\left(x^{n}\right)=n x^{n-1}
\]

Step 3: Substitute \(n=\tfrac{1}{2}\)

Apply the rule to \(x^{1/2}\):

\[
\frac{d}{dx}\left(x^{1/2}\right)=\frac{1}{2}x^{\,1/2-1}
\]

Step 4: Simplify the exponent

Compute the exponent \( \tfrac{1}{2}-1\):

\[
\frac{1}{2}-1=\frac{1}{2}-\frac{2}{2}=-\frac{1}{2}
\]

Step 5: Write the final derivative

Substitute back:

\[
\frac{d}{dx}\left(x^{1/2}\right)=\frac{1}{2}x^{-1/2}
\]

Optional Step: Rewrite with a positive exponent

Since \(x^{-1/2}=\dfrac{1}{\sqrt{x}}\), we can also write:

\[
\frac{d}{dx}\left(x^{1/2}\right)=\frac{1}{2\sqrt{x}}
\]

Answer: \(\displaystyle \frac{d}{dx}\left(x^{1/2}\right)=\frac{1}{2}x^{-1/2}=\frac{1}{2\sqrt{x}}\).

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Calculus FAQ

What is the derivative of \(x^{1/2}\)?

Use the power rule: \( \dfrac{d}{dx}x^{n}=nx^{n-1}\). Here \(n=\tfrac{1}{2}\), so \(\dfrac{d}{dx}x^{1/2}=\tfrac{1}{2}x^{-1/2}=\dfrac{1}{2\sqrt{x}}\), for \(x>0\).

How do you rewrite \(\sqrt{x}\) to differentiate it?

Rewrite \(x^{1/2}\). Then apply the power rule: \(\dfrac{d}{dx}\sqrt{x}=\dfrac{d}{dx}x^{1/2}=\tfrac{1}{2}x^{-1/2}\).

What is the derivative of \(\sqrt{x}\) using limits/definition?

Start with \(f(x)=x^{1/2}\). Compute \(f'(x)=\lim_{h\to 0}\dfrac{(x+h)^{1/2}-x^{1/2}}{h}\). Rationalize, simplify, and get \(f'(x)=\tfrac{1}{2}x^{-1/2}\).

What is the domain of the derivative for \(x^{1/2}\)?

Since \(\sqrt{x}\) is real only for \(x\ge 0\), the derivative formula \(\tfrac{1}{2}x^{-1/2}\) is valid for \(x>0\). At \(x=0\), it is undefined/infinite.

What is the derivative of \(x^{1/2}\) for negative \(x\) (real-valued)?

For real calculus, \(\sqrt{x}\) is not defined when \(x<0\). So the derivative \(\tfrac{1}{2}x^{-1/2}\) is not real-valued there.

How to differentiate \(x^{1/2}\) using the chain rule?

If written as \(\sqrt{u}\) with \(u=x\), then \(\dfrac{d}{dx}\sqrt{u}=\dfrac{1}{2\sqrt{u}}\cdot \dfrac{du}{dx}\). With \(u=x\), this gives \(\tfrac{1}{2\sqrt{x}}\).

Is \(\dfrac{d}{dx}x^{1/2}=\dfrac{1}{2\sqrt{x}}\) always correct?

It is correct for \(x>0\). More generally, \(\tfrac{1}{2}x^{-1/2}\) matches the same derivative wherever \(x^{1/2}\) is differentiable in the chosen real domain.

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