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

Q: What is the derivative of \frac{1}{x^3} ?

Rewrite \frac{1}{x^3}=x^{-3}. Then \frac{d}{dx}x^{-3}=-3x^{-4}=-\frac{3}{x^4}.

Q: How do you use the power rule on x^{-3}?

Power rule: \frac{d}{dx}x^n=nx^{n-1}. With n=-3: \frac{d}{dx}x^{-3}=-3x^{-4}.

Q: What is the derivative of x^{-3} directly?

Apply nx^{n-1}: \frac{d}{dx}x^{-3}=-3x^{-4}. This equals -\frac{3}{x^4} for x\neq 0.

Q: Can you differentiate using the quotient rule for \frac{1}{x^3} ?

Use f(x)=1, g(x)=x^3. Then \left(\frac{f}{g}\right)'=\frac{f'g-fg'}{g^2}=\frac{0\cdot x^3-1\cdot 3x^2}{(x^3)^2}=-\frac{3x^2}{x^6}=-\frac{3}{x^4}.

Q: What is \frac{d}{dx}\left(\frac{1}{x^3}\right) in terms of negative powers?

\frac{1}{x^3}=x^{-3}, so the derivative is -3x^{-4}.

Q: What is the derivative for \frac{1}{(x^3)} and is there any domain issue?

The derivative is -\frac{3}{x^4}, valid for x\neq 0. The original function \frac{1}{x^3} is undefined at x=0.

Q: What is the second derivative of \frac{1}{x^3}?

First derivative -3x^{-4}. Then differentiate again: (-3)\cdot(-4)x^{-5}=12x^{-5}=\frac{12}{x^5}.

Answer

Rewrite \( \frac{1}{x^3} \) as \( x^{-3} \). Differentiate using the power rule \( \frac{d}{dx}x^n = nx^{n-1} \).

\[
\frac{d}{dx}\left(x^{-3}\right)= -3x^{-4} = -\frac{3}{x^4}
\]

Final result: \( -\frac{3}{x^4} \).

Detailed Explanation

We want to find the derivative of the function

\[
f(x)=\frac{1}{x^3}.
\]

Step 1: Rewrite the function using exponents.

Fraction powers are easier to differentiate when written as powers of \(x\):

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

Step 2: Use the power rule for derivatives.

The power rule says: if \(f(x)=x^n\), then

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

Step 3: Differentiate \(x^{-3}\).

Here, \(n=-3\). So

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

Combine the exponents:

\[
(-3)x^{-4}.
\]

Step 4: Rewrite with a positive exponent in the denominator (optional).

Since \(x^{-4}=\frac{1}{x^4}\), we get

\[
(-3)x^{-4}=-\frac{3}{x^4}.
\]

Final Answer

\[
\frac{d}{dx}\left(\frac{1}{x^3}\right)=-\frac{3}{x^4}.
\]

See full solution

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Calculus FAQ

What is the derivative of \( \frac{1}{x^3} \)?

Rewrite \( \frac{1}{x^3}=x^{-3}\). Then \( \frac{d}{dx}x^{-3}=-3x^{-4}=-\frac{3}{x^4}\).

How do you use the power rule on \(x^{-3}\)?

Power rule: \( \frac{d}{dx}x^n=nx^{n-1}\). With \(n=-3\): \( \frac{d}{dx}x^{-3}=-3x^{-4}\).

What is the derivative of \(x^{-3}\) directly?

Apply \(nx^{n-1}\): \( \frac{d}{dx}x^{-3}=-3x^{-4}\). This equals \( -\frac{3}{x^4}\) for \(x\neq 0\).

Can you differentiate using the quotient rule for \( \frac{1}{x^3} \)?

Use \(f(x)=1\), \(g(x)=x^3\). Then \( \left(\frac{f}{g}\right)'=\frac{f'g-fg'}{g^2}=\frac{0\cdot x^3-1\cdot 3x^2}{(x^3)^2}=-\frac{3x^2}{x^6}=-\frac{3}{x^4}\).

What is \( \frac{d}{dx}\left(\frac{1}{x^3}\right)\) in terms of negative powers?

\( \frac{1}{x^3}=x^{-3}\), so the derivative is \( -3x^{-4}\).

What is the derivative for \( \frac{1}{(x^3)}\) and is there any domain issue?

The derivative is \( -\frac{3}{x^4}\), valid for \(x\neq 0\). The original function \( \frac{1}{x^3}\) is undefined at \(x=0\).

What is the second derivative of \( \frac{1}{x^3}\)?

First derivative \( -3x^{-4}\). Then differentiate again: \( (-3)\cdot(-4)x^{-5}=12x^{-5}=\frac{12}{x^5}\).

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