Q. \[\frac{d}{dx}\left(\frac{1}{\sqrt{x}}\right)\]

Q: What is the derivative of \frac{1}{\sqrt{x}} ?

Rewrite \frac{1}{\sqrt{x}} = x^{-1/2} . Then \frac{d}{dx}x^{-1/2} = -\frac{1}{2}x^{-3/2} = -\frac{1}{2x^{3/2}} .

Q: How do you differentiate using the chain rule for x^{-1/2} ?

Use \frac{d}{dx}[u(x)]^n = n[u(x)]^{n-1}u'(x) . Here u=x , n=-\frac{1}{2} . So -\frac{1}{2}x^{-3/2} .

Q: What is the derivative of x^{-\frac{1}{2}} directly?

Apply the power rule: \frac{d}{dx}x^a = ax^{a-1} . With a=-\frac{1}{2} , derivative is -\frac{1}{2}x^{-3/2} .

Q: What is the derivative of \frac{1}{\sqrt{x}} for x>0 ?

For x>0 , the derivative is -\frac{1}{2x^{3/2}} . The function and derivative are real-valued on this domain.

Answer

Rewrite

\[
\frac{1}{\sqrt{x}}=x^{-\frac{1}{2}}.
\]

Differentiate using the power rule \( \frac{d}{dx}x^n=nx^{n-1} \).

\[
\frac{d}{dx}\left(x^{-\frac{1}{2}}\right)=-\frac{1}{2}x^{-\frac{3}{2}}=-\frac{1}{2x^{3/2}}.
\]

Final result:

\[
-\frac{1}{2x^{3/2}}.
\]

Detailed Explanation

We want to find the derivative of the function

\[
f(x)=\frac{1}{\sqrt{x}}.
\]

Step 1: Rewrite using exponents

Recall that

\[
\sqrt{x}=x^{1/2}.
\]

Therefore,

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

Step 2: Apply the power rule

The power rule says: if

\[
g(x)=x^{n},
\]

then

\[
g'(x)=n x^{n-1}.
\]

Here, \(n=-\frac{1}{2}\). So differentiate \(x^{-1/2}\):

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

Step 3: Simplify the exponent

Compute the exponent:

\[
-\frac{1}{2}-1=-\frac{1}{2}-\frac{2}{2}=-\frac{3}{2}.
\]

So the derivative becomes:

\[
f'(x)=-\frac{1}{2}x^{-3/2}.
\]

Step 4: Rewrite in radical form (optional)

Convert \(x^{-3/2}\) back to a denominator with a square root:

\[
x^{-3/2}=\frac{1}{x^{3/2}}=\frac{1}{x\sqrt{x}}.
\]

\[
f'(x)=-\frac{1}{2}\cdot \frac{1}{x^{3/2}}=-\frac{1}{2x^{3/2}}=-\frac{1}{2x\sqrt{x}}.
\]

Final Answer

\[
\frac{d}{dx}\left(\frac{1}{\sqrt{x}}\right)=-\frac{1}{2x^{3/2}}=-\frac{1}{2x\sqrt{x}}.
\]

See full solution

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Calculus FAQ

What is the derivative of \( \frac{1}{\sqrt{x}} \) ?

Rewrite \( \frac{1}{\sqrt{x}} = x^{-1/2} \). Then \( \frac{d}{dx}x^{-1/2} = -\frac{1}{2}x^{-3/2} = -\frac{1}{2x^{3/2}} \).

How do you differentiate using the chain rule for \( x^{-1/2} \)?

Use \( \frac{d}{dx}[u(x)]^n = n[u(x)]^{n-1}u'(x) \). Here \( u=x \), \( n=-\frac{1}{2} \). So \( -\frac{1}{2}x^{-3/2} \).

What is the derivative of \( x^{-\frac{1}{2}} \) directly?

Apply the power rule: \( \frac{d}{dx}x^a = ax^{a-1} \). With \( a=-\frac{1}{2} \), derivative is \( -\frac{1}{2}x^{-3/2} \).

Can you express the derivative with radicals (no negative exponents)?

\( -\frac{1}{2}x^{-3/2} = -\frac{1}{2\left(x^{3/2}\right)} = -\frac{1}{2x\sqrt{x}} \).

What is the derivative of \( \frac{1}{\sqrt{x}} \) for \( x>0 \)?

For \( x>0 \), the derivative is \( -\frac{1}{2x^{3/2}} \). The function and derivative are real-valued on this domain.

What happens if the function is \( \sqrt{x} \) instead of \( \frac{1}{\sqrt{x}} \)?

\( \frac{d}{dx}\sqrt{x} = \frac{d}{dx}x^{1/2} = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}} \).

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