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

Q: What is the derivative of \ln(x^2) (real x)?

Use \ln(x^2)=2\ln|x|. Then \dfrac{d}{dx}\ln|x|=\dfrac{1}{x} for x\ne 0. So \dfrac{d}{dx}\ln(x^2)=\dfrac{2}{x}, for x\ne 0.

Q: Does the chain rule give the same result?

Let u=x^2. Then \dfrac{d}{dx}\ln(u)=\dfrac{u'}{u}. Here u'=2x, so \dfrac{d}{dx}\ln(x^2)=\dfrac{2x}{x^2}=\dfrac{2}{x} for x\ne 0.

Q: What about the domain—where is \ln(x^2) defined?

Need x^2>0 for the log input to be positive, and also x\ne 0. So \ln(x^2) is defined for all real x\ne 0.

Q: What happens if x>0?

For x>0, |x|=x, so \ln(x^2)=2\ln x. Then \dfrac{d}{dx}2\ln x=\dfrac{2}{x} for x>0.

Q: What if x<0?

For x<0, |x|=-x, so \ln(x^2)=2\ln|x|. Since \dfrac{d}{dx}\ln|x|=\dfrac{1}{x} for x\ne 0, the derivative is still \dfrac{2}{x} for x<0.

Q: Can we differentiate \ln(x^2) incorrectly by writing \ln(x)^2?

No. \ln(x^2)\ne (\ln x)^2. The first is a log of a square, the second is square of a log. Using the wrong expression leads to a different derivative.

Answer

To find the derivative of \( \ln(x^2) \), use the chain rule.

\[
\frac{d}{dx}\ln(x^2)=\frac{1}{x^2}\cdot \frac{d}{dx}(x^2)=\frac{1}{x^2}\cdot 2x=\frac{2}{x}.
\]

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

Detailed Explanation

We want to find the derivative of the function \( \ln(x^2) \).

Step 1: Identify the outer and inner functions

The function has the form \( \ln(u) \) where \( u = x^2 \).

Step 2: Use the chain rule

The derivative of \( \ln(u) \) with respect to \( x \) is

\[
\frac{d}{dx}\bigl(\ln(u)\bigr) = \frac{1}{u}\cdot \frac{du}{dx}.
\]

Here, \( u = x^2 \), so we need \( \frac{du}{dx} \).

Step 3: Differentiate the inner function

Compute \( \frac{d}{dx}(x^2) \).

\[
\frac{du}{dx}=\frac{d}{dx}(x^2)=2x.
\]

Step 4: Substitute into the chain rule

Now plug \( u = x^2 \) and \( \frac{du}{dx}=2x \) into the formula.

\[
\frac{d}{dx}\bigl(\ln(x^2)\bigr)=\frac{1}{x^2}\cdot 2x.
\]

Step 5: Simplify

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

Final Answer

\[
\frac{d}{dx}\bigl(\ln(x^2)\bigr)=\frac{2}{x},
\]
for all \( x \neq 0 \) (since \( \ln(x^2) \) is defined only when \( x^2>0 \), which means \( x \neq 0 \)).

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Calculus FAQ

What is the derivative of \( \ln(x^2)\) (real \(x\))?

Use \( \ln(x^2)=2\ln|x|\). Then \( \dfrac{d}{dx}\ln|x|=\dfrac{1}{x}\) for \(x\ne 0\). So \( \dfrac{d}{dx}\ln(x^2)=\dfrac{2}{x}\), for \(x\ne 0\).

Does the chain rule give the same result?

Let \(u=x^2\). Then \( \dfrac{d}{dx}\ln(u)=\dfrac{u'}{u}\). Here \(u'=2x\), so \( \dfrac{d}{dx}\ln(x^2)=\dfrac{2x}{x^2}=\dfrac{2}{x}\) for \(x\ne 0\).

What about the domain—where is \( \ln(x^2)\) defined?

Need \(x^2>0\) for the log input to be positive, and also \(x\ne 0\). So \( \ln(x^2)\) is defined for all real \(x\ne 0\).

What happens if \(x>0\)?

For \(x>0\), \( |x|=x\), so \( \ln(x^2)=2\ln x\). Then \( \dfrac{d}{dx}2\ln x=\dfrac{2}{x}\) for \(x>0\).

What if \(x<0\)?

For \(x<0\), \( |x|=-x\), so \( \ln(x^2)=2\ln|x|\). Since \( \dfrac{d}{dx}\ln|x|=\dfrac{1}{x}\) for \(x\ne 0\), the derivative is still \( \dfrac{2}{x}\) for \(x<0\).

Can we differentiate \( \ln(x^2)\) incorrectly by writing \( \ln(x)^2\)?

No. \( \ln(x^2)\ne (\ln x)^2\). The first is a log of a square, the second is square of a log. Using the wrong expression leads to a different derivative.

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