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

Q: What is \frac{d}{dx}\left(\frac{1}{x^2}\right) ?

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

Q: Using the power rule, what is \frac{d}{dx}\left(x^{-2}\right) ?

Power rule: \frac{d}{dx}(x^n)=nx^{n-1} . With n=-2 , \frac{d}{dx}(x^{-2})=-2x^{-3} .

Q: Can I use the quotient rule to differentiate \frac{1}{x^2} ?

Yes. Let f=\frac{1}{x^2}=\frac{1}{x^2} as quotient \frac{1}{x^2} . Result is f'(x)=-\frac{2}{x^3} .

Q: What is the derivative of x^{-2} written as a fraction?

Start with \frac{d}{dx}(x^{-2})=-2x^{-3} . Convert to fraction: -2x^{-3}=-\frac{2}{x^3} .

Q: Where is the derivative undefined for \frac{1}{x^2} ?

\frac{1}{x^2} and -\frac{2}{x^3} are undefined at x=0 . So the derivative does not exist at x=0 .

Q: What is \frac{d}{dx}\left(\frac{1}{x^2}\right) written with exponent notation?

With exponent notation, \frac{1}{x^2}=x^{-2} . The derivative is -2x^{-3} .

Q: Differentiate \frac{1}{x^2} using the chain rule.

Let u=x^2 . Then \frac{1}{x^2}=u^{-1} . So \frac{d}{dx}(u^{-1})=-u^{-2}\cdot u'=-x^{-4}\cdot 2x=-\frac{2}{x^3} .

Answer

We rewrite \( \frac{1}{x^2} \) as \(x^{-2}\). Using the power rule \( \frac{d}{dx}x^n = nx^{n-1}\), we get

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

Final result: \(-\frac{2}{x^3}\).

Detailed Explanation

We want to find the derivative of the function

\[
f(x)=\frac{1}{x^2}.
\]

Step 1: Rewrite the function using exponents.

Recall that

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

So we can rewrite the function as

\[
f(x)=x^{-2}.
\]

Step 2: Use the power rule for derivatives.

The power rule says:

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

Here, \(n=-2\). Substitute \(n=-2\) into the power rule.

Step 3: Differentiate.

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

Step 4: Rewrite the result in a more standard fraction form.

Recall that \(x^{-3}=\frac{1}{x^3}\). Therefore:

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

Final Answer

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

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Calculus FAQ

What is \( \frac{d}{dx}\left(\frac{1}{x^2}\right) \)?

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

Using the power rule, what is \( \frac{d}{dx}\left(x^{-2}\right) \)?

Power rule: \( \frac{d}{dx}(x^n)=nx^{n-1} \). With \( n=-2 \), \( \frac{d}{dx}(x^{-2})=-2x^{-3} \).

Can I use the quotient rule to differentiate \( \frac{1}{x^2} \)?

Yes. Let \( f=\frac{1}{x^2}=\frac{1}{x^2} \) as quotient \( \frac{1}{x^2} \). Result is \( f'(x)=-\frac{2}{x^3} \).

What is the derivative of \( x^{-2} \) written as a fraction?

Start with \( \frac{d}{dx}(x^{-2})=-2x^{-3} \). Convert to fraction: \( -2x^{-3}=-\frac{2}{x^3} \).

Where is the derivative undefined for \( \frac{1}{x^2} \)?

\( \frac{1}{x^2} \) and \( -\frac{2}{x^3} \) are undefined at \( x=0 \). So the derivative does not exist at \( x=0 \).

What is \( \frac{d}{dx}\left(\frac{1}{x^2}\right) \) written with exponent notation?

With exponent notation, \( \frac{1}{x^2}=x^{-2} \). The derivative is \( -2x^{-3} \).

Differentiate \( \frac{1}{x^2} \) using the chain rule.

Let \( u=x^2 \). Then \( \frac{1}{x^2}=u^{-1} \). So \( \frac{d}{dx}(u^{-1})=-u^{-2}\cdot u'=-x^{-4}\cdot 2x=-\frac{2}{x^3} \).

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