Q. \[ \frac{d}{dx}\left(4^x\right) \]

Q: What is the derivative of 4^x?

Use \frac{d}{dx}\big(a^x\big)=a^x\ln(a) for a>0. So \frac{d}{dx}\big(4^x\big)=4^x\ln(4).

Q: Why does the natural logarithm \ln(4) appear?

Because a^x=e^{x\ln(a)}. Differentiate: \frac{d}{dx}e^{x\ln(a)}=e^{x\ln(a)}\ln(a)=a^x\ln(a). For a=4, it becomes 4^x\ln(4).

Q: How do you differentiate 4^{x+2}?

Rewrite 4^{x+2}=4^2\cdot 4^x. Then \frac{d}{dx}(4^{x+2})=4^2\cdot 4^x\ln(4)=16\cdot 4^x\ln(4).

Q: What is the derivative of 4^{3x}?

Use \frac{d}{dx}(a^{kx})=a^{kx}\ln(a)\cdot k. So \frac{d}{dx}(4^{3x})=4^{3x}\ln(4)\cdot 3=3\ln(4)\,4^{3x}.

Q: What is the derivative of 4^{x^2}?

Let y=4^{x^2}=e^{x^2\ln(4)}. Then \frac{d}{dx}(4^{x^2})=e^{x^2\ln(4)}\cdot (2x\ln(4))=2x\ln(4)\,4^{x^2}.

Q: Can you differentiate 4^{-x}?

Use 4^{-x}=(4^{-1})^x or apply the formula directly: \frac{d}{dx}(a^{-x})=a^{-x}\ln(a)\cdot(-1). Thus \frac{d}{dx}(4^{-x})=-4^{-x}\ln(4).

Answer

Rewrite \(4^x\) using the exponential form: \(4^x=e^{x\ln 4}\). Differentiate:

\[
\frac{d}{dx}\left(4^x\right)=\frac{d}{dx}\left(e^{x\ln 4}\right)=e^{x\ln 4}\cdot \ln 4=4^x\ln 4
\]

Final result: \(\frac{d}{dx}\left(4^x\right)=4^x\ln 4\).

Detailed Explanation

We want to find the derivative of the function \(f(x)=4^x\).

Step 1: Identify the base and the exponential form

The function is of the form \(a^x\), where \(a=4\). We need the derivative rule for exponential functions with a constant base.

Step 2: Use the exponential derivative rule

A standard rule is:

If \(f(x)=a^x\) where \(a>0\) and \(a\neq 1\), then

\(\displaystyle \frac{d}{dx}\left(a^x\right)=a^x\ln(a)\).

Step 3: Substitute \(a=4\) into the rule

Substitute \(a=4\) into the formula:

\(\displaystyle \frac{d}{dx}\left(4^x\right)=4^x\ln(4)\).

Step 4: State the final answer

The derivative of \(4^x\) is:

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

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Calculus FAQ

What is the derivative of \(4^x\)?

Use \( \frac{d}{dx}\big(a^x\big)=a^x\ln(a)\) for \(a>0\). So \(\frac{d}{dx}\big(4^x\big)=4^x\ln(4)\).

Why does the natural logarithm \(\ln(4)\) appear?

Because \(a^x=e^{x\ln(a)}\). Differentiate: \(\frac{d}{dx}e^{x\ln(a)}=e^{x\ln(a)}\ln(a)=a^x\ln(a)\). For \(a=4\), it becomes \(4^x\ln(4)\).

How do you differentiate \(4^{x+2}\)?

Rewrite \(4^{x+2}=4^2\cdot 4^x\). Then \(\frac{d}{dx}(4^{x+2})=4^2\cdot 4^x\ln(4)=16\cdot 4^x\ln(4)\).

What is the derivative of \(4^{3x}\)?

Use \( \frac{d}{dx}(a^{kx})=a^{kx}\ln(a)\cdot k\). So \(\frac{d}{dx}(4^{3x})=4^{3x}\ln(4)\cdot 3=3\ln(4)\,4^{3x}\).

What is the derivative of \(4^{x^2}\)?

Let \(y=4^{x^2}=e^{x^2\ln(4)}\). Then \(\frac{d}{dx}(4^{x^2})=e^{x^2\ln(4)}\cdot (2x\ln(4))=2x\ln(4)\,4^{x^2}\).

Can you differentiate \(4^{-x}\)?

Use \(4^{-x}=(4^{-1})^x\) or apply the formula directly: \(\frac{d}{dx}(a^{-x})=a^{-x}\ln(a)\cdot(-1)\). Thus \(\frac{d}{dx}(4^{-x})=-4^{-x}\ln(4)\).

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