Q. \[ \frac{d}{dx} \left( 7^{x} \right) \]

Q: How do you differentiate (7^x)^5?

(7^x)^5 = 7^{5x}, so \frac{d}{dx}7^{5x} = 7^{5x}\ln(7)\cdot 5.

Q: What is \frac{d}{dx}(7^{x+1})?

Use shift: 7^{x+1} derivative is 7^{x+1}\ln(7).

Q: What is \frac{d}{dx}(7^{-x})?

\frac{d}{dx}7^{-x} = 7^{-x}\ln(7)\cdot(-1) = -7^{-x}\ln(7).

Q: How do you differentiate \frac{d}{dx}\left(3\cdot 7^x\right)?

Constant multiple rule: \frac{d}{dx}\left(3\cdot 7^x\right)=3\cdot 7^x\ln(7).

Q: Why does a \ln(7) appear in the derivative?

Because a^x = e^{x\ln(a)}, and \frac{d}{dx}e^{u}=e^{u}u'. Here u=x\ln(7), so u'=\ln(7).

Answer

To differentiate \(7^x\), use the rule \( \frac{d}{dx} a^x = a^x \ln(a)\) for \(a>0\), \(a\ne1\).

\[
\frac{d}{dx}\left(7^x\right)=7^x\ln(7).
\]

Detailed Explanation

We want to find the derivative of the function \(f(x)=7^x\).

Step 1: Identify the function type.

The base \(7\) is a constant, and the exponent is the variable \(x\). So this is an exponential function of the form \(a^x\) where \(a=7\).

Step 2: Use the exponential derivative rule.

A key rule is:

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

This works for any constant \(a>0\) with \(a\ne 1\).

Step 3: Substitute \(a=7\).

Apply the rule to \(7^x\):

\[
\frac{d}{dx}\left(7^x\right)=7^x\ln(7)
\]

Final Answer.

\[
\boxed{\frac{d}{dx}\left(7^x\right)=7^x\ln(7)}
\]

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Calculus FAQ

What is \(\frac{d}{dx}7^x\)?

\(\frac{d}{dx}7^x = 7^x\ln(7)\).

What is the derivative of \(a^x\) for constant \(a>0\), \(a\neq 1\)?

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

How do you differentiate \((7^x)^5\)?

\((7^x)^5 = 7^{5x}\), so \(\frac{d}{dx}7^{5x} = 7^{5x}\ln(7)\cdot 5\).

What is \(\frac{d}{dx}(7^{x+1})\)?

Use shift: \(7^{x+1}\) derivative is \(7^{x+1}\ln(7)\).

What is \(\frac{d}{dx}(7^{-x})\)?

\(\frac{d}{dx}7^{-x} = 7^{-x}\ln(7)\cdot(-1) = -7^{-x}\ln(7)\).

How do you differentiate \(\frac{d}{dx}\left(3\cdot 7^x\right)\)?

Constant multiple rule: \(\frac{d}{dx}\left(3\cdot 7^x\right)=3\cdot 7^x\ln(7)\).

Why does a \(\ln(7)\) appear in the derivative?

Because \(a^x = e^{x\ln(a)}\), and \(\frac{d}{dx}e^{u}=e^{u}u'\). Here \(u=x\ln(7)\), so \(u'=\ln(7)\).

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