Q. \[ \text{Find the derivative of } 2^x. \]

Q: What is the derivative of 2^x?

\\\dfrac{d}{dx}\\big(2^x\\big)=2^x\\ln 2\.\

Q: Why does d(a^x)/dx=a^x\\ln(a) work?

Because a^x=e^{x\\ln a}. Then \\dfrac{d}{dx}e^{x\\ln a}=e^{x\\ln a}(\\ln a)=a^x\\ln a\.

Q: What is the derivative of 5^{x}?

\\\dfrac{d}{dx}(5^x)=5^x\\ln 5\.\

Q: What is \\dfrac{d}{dx}(2^{3x})\?

Use chain rule: \2^{3x}=e^{3x\\ln 2}\. So \\\dfrac{d}{dx}(2^{3x})=2^{3x}\\ln 2\\cdot 3\.\

Q: What is \\dfrac{d}{dx}(2^{x+1})\?

\2^{x+1}=2\\cdot 2^x\. Thus \\\dfrac{d}{dx}(2^{x+1})=2\\cdot 2^x\\ln 2=2^{x+1}\\ln 2\.\

Q: Can you differentiate 2^{-x}?

2^{-x}=(2^x)^{-1}, or write 2^{-x}=e^{-x\\ln 2}\. Then \\\dfrac{d}{dx}(2^{-x})=-2^{-x}\\ln 2\.\

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

\\\ln(2^x)=x\\ln 2\. So \\\dfrac{d}{dx}\\ln(2^x)=\\ln 2\.\

Answer

To differentiate \(2^x\), use the rule:

\( \dfrac{d}{dx} a^x = a^x \ln(a) \) for \(a>0\), \(a\neq 1\).

Here \(a=2\), so

\[
\dfrac{d}{dx} 2^x = 2^x \ln(2)
\]

Final result: \(2^x \ln(2)\).

Detailed Explanation

We want to find the derivative of \(2^x\) with respect to \(x\).

Step 1: Recognize the function form

The function is an exponential with a constant base: \(\,2^x\). For derivatives of the form \(a^x\) (where \(a\) is a positive constant and \(a \ne 1\)), we use the rule:

\[
\frac{d}{dx}\left(a^x\right)=a^x\ln(a)
\]

Here, \(a=2\).

Step 2: Apply the rule with \(a=2\)

Substitute \(a=2\) into the formula:

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

Final Answer

\[
\boxed{\frac{d}{dx}\left(2^x\right)=2^x\ln(2)}
\]

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Calculus FAQ

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

\(\\(\\dfrac{d}{dx}\\big(2^x\\big)=2^x\\ln 2\\).\\)

Why does \(d(a^x)/dx=a^x\\ln(a)\) work?

Because \(a^x=e^{x\\ln a}\). Then \(\\dfrac{d}{dx}e^{x\\ln a}=e^{x\\ln a}(\\ln a)=a^x\\ln a\\).

What is the derivative of \(5^{x}\)?

\(\\(\\dfrac{d}{dx}(5^x)=5^x\\ln 5\\).\\)

What is \(\\dfrac{d}{dx}(2^{3x})\\)?

Use chain rule: \(\\(2^{3x}=e^{3x\\ln 2}\\). So \(\\(\\dfrac{d}{dx}(2^{3x})=2^{3x}\\ln 2\\cdot 3\\).\\)

What is \(\\dfrac{d}{dx}(2^{x+1})\\)?

\(\\(2^{x+1}=2\\cdot 2^x\\). Thus \(\\(\\dfrac{d}{dx}(2^{x+1})=2\\cdot 2^x\\ln 2=2^{x+1}\\ln 2\\).\\)

Can you differentiate \(2^{-x}\)?

\(2^{-x}=(2^x)^{-1}\), or write \(2^{-x}=e^{-x\\ln 2}\\). Then \(\\(\\dfrac{d}{dx}(2^{-x})=-2^{-x}\\ln 2\\).\\)

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

\(\\(\\ln(2^x)=x\\ln 2\\). So \(\\(\\dfrac{d}{dx}\\ln(2^x)=\\ln 2\\).\\)

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