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

Q: What is the derivative of x^e where e is a constant?

Use the power rule: \frac{d}{dx}x^e = e x^{e-1}.

Q: Does the rule \frac{d}{dx}x^n = nx^{n-1} apply when n=e?

Yes, if e is treated as a constant exponent. Then \frac{d}{dx}x^e = e x^{e-1}.

Q: How do you derive \frac{d}{dx}x^e using logarithmic differentiation?

Let y=x^e. Take logs: \ln y = e\ln x. Differentiate: \frac{y'}{y} = \frac{e}{x}. So y' = e x^{e-1}.

Q: What is \frac{d}{dx}\big(x^{e}\big) for x>0?

For real-valued x^e, assume x>0. Then \frac{d}{dx}x^e = e x^{e-1}.

Q: What is the domain where \frac{d}{dx}x^e is real if e is not restricted to an integer?

For real exponents e, x^e is real for x>0. On that domain, the derivative is e x^{e-1}.

Q: Is \frac{d}{dx}x^e the same as \frac{d}{dx}e^x?

No. \frac{d}{dx}x^e = e x^{e-1}, while \frac{d}{dx}e^x = e^x.

Answer

To differentiate \(x^e\) (where \(e\) is a constant), use the power rule. For \(x^n\), the derivative is \(n x^{n-1}\). Here \(n=e\), so:

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

Detailed Explanation

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

Step 1: Identify the constant and the variable

Here, \(x\) is the variable. The exponent \(e\) is a constant (since \(e \approx 2.718\ldots\) but, importantly, it does not depend on \(x\)). So the function has the form

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

Step 2: Recall the power rule

The power rule says: if \(n\) is a constant, then

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

Step 3: Apply the power rule with \(n=e\)

Substitute \(n=e\) into the power rule:

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

Final answer

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

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Calculus FAQ

What is the derivative of \(x^e\) where \(e\) is a constant?

Use the power rule: \(\frac{d}{dx}x^e = e x^{e-1}\).

Does the rule \(\frac{d}{dx}x^n = nx^{n-1}\) apply when \(n=e\)?

Yes, if \(e\) is treated as a constant exponent. Then \(\frac{d}{dx}x^e = e x^{e-1}\).

How do you derive \(\frac{d}{dx}x^e\) using logarithmic differentiation?

Let \(y=x^e\). Take logs: \(\ln y = e\ln x\). Differentiate: \(\frac{y'}{y} = \frac{e}{x}\). So \(y' = e x^{e-1}\).

What is \(\frac{d}{dx}\big(x^{e}\big)\) for \(x>0\)?

For real-valued \(x^e\), assume \(x>0\). Then \(\frac{d}{dx}x^e = e x^{e-1}\).

What is the domain where \(\frac{d}{dx}x^e\) is real if \(e\) is not restricted to an integer?

For real exponents \(e\), \(x^e\) is real for \(x>0\). On that domain, the derivative is \(e x^{e-1}\).

Is \(\frac{d}{dx}x^e\) the same as \(\frac{d}{dx}e^x\)?

No. \(\frac{d}{dx}x^e = e x^{e-1}\), while \(\frac{d}{dx}e^x = e^x\).

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