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

Q: Why does \ln(3) appear in the derivative?

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

Q: What is the derivative of \left(3^{x}\right)^2 ?

\left(3^x\right)^2=3^{2x}. So \frac{d}{dx}(3^{2x})=3^{2x}\ln(3)\cdot 2=2\ln(3)\,3^{2x}.

Q: What is \frac{d}{dx}\left(3^{3x}\right) ?

Use \frac{d}{dx}(a^{u})=a^{u}\ln(a)\,u'. Here u=3x, so u'=3. Thus \frac{d}{dx}(3^{3x})=3\ln(3)\,3^{3x}.

Q: What is \frac{d}{dx}\left(3^{x+1}\right) ?

Rewrite 3^{x+1}=3\cdot 3^x. Then \frac{d}{dx}(3\cdot 3^x)=3\cdot 3^x\ln(3)=3^{x+1}\ln(3).

Q: What is \frac{d}{dx}\left(3^{-x}\right) ?

3^{-x}=a^{u} with a=3, u=-x. So \frac{d}{dx}(3^{-x})=3^{-x}\ln(3)\cdot(-1)=-\ln(3)\,3^{-x}.

Q: What is the general rule for \frac{d}{dx}(a^x) ?

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

Answer

To find the derivative of \(3^x\), use the rule \(\frac{d}{dx}\left(a^x\right)=a^x\ln(a)\) for \(a>0\).

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

Detailed Explanation

We want to find the derivative of the function \(y = 3^x\).

Step 1: Identify the function type

The function \(3^x\) is an exponential function where the base \(3\) is a constant and the exponent is the variable \(x\).

Step 2: Use the general differentiation rule

For any positive constant \(a \neq 1\), the derivative is:

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

Here, the base \(a\) is \(3\).

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

Replace \(a\) with \(3\):

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

Step 4: Final answer

Therefore, the derivative of \(3^x\) is:

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

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Calculus FAQ

What is \( \frac{d}{dx}\left(3^x\right) \)?

\(\frac{d}{dx}(3^x)=3^x\ln(3)\).

Why does \( \ln(3) \) appear in the derivative?

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

What is the derivative of \( \left(3^{x}\right)^2 \)?

\(\left(3^x\right)^2=3^{2x}\). So \(\frac{d}{dx}(3^{2x})=3^{2x}\ln(3)\cdot 2=2\ln(3)\,3^{2x}\).

What is \( \frac{d}{dx}\left(3^{3x}\right) \)?

Use \(\frac{d}{dx}(a^{u})=a^{u}\ln(a)\,u'\). Here \(u=3x\), so \(u'=3\). Thus \(\frac{d}{dx}(3^{3x})=3\ln(3)\,3^{3x}\).

What is \( \frac{d}{dx}\left(3^{x+1}\right) \)?

Rewrite \(3^{x+1}=3\cdot 3^x\). Then \(\frac{d}{dx}(3\cdot 3^x)=3\cdot 3^x\ln(3)=3^{x+1}\ln(3)\).

What is \( \frac{d}{dx}\left(3^{-x}\right) \)?

\(3^{-x}=a^{u}\) with \(a=3\), \(u=-x\). So \(\frac{d}{dx}(3^{-x})=3^{-x}\ln(3)\cdot(-1)=-\ln(3)\,3^{-x}\).

What is the general rule for \( \frac{d}{dx}(a^x) \)?

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

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