Question:medium

If $f(x) = \begin{cases} 4(5^{x}) & x < 0 \\ 8k + x & x \ge 0 \end{cases}$ then $f'(-1) =$

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Always identify which "piece" of a piecewise function your point falls into before differentiating.
  • $\frac{2}{5}\log 5$
  • $\frac{4}{5}\log 5$
  • $\frac{3}{5}\log 5$
  • $20\log 5$
Show Solution

The Correct Option is B

Solution and Explanation

Step 1: Understanding the Concept:
To find the derivative of a piecewise function at a specific point, we use the rule defined for the interval containing that point. Since we need \( f'(-1) \) and \( -1<0 \), we only consider the piece \( f(x) = 4(5^x) \)[cite: 1].
Step 2: Key Formula or Approach:
The derivative of an exponential function \( a^x \) is \( \frac{d}{dx}(a^x) = a^x \log a \) (or \( a^x \ln a \))[cite: 1].
Step 3: Detailed Explanation:
For \( x<0 \), \( f(x) = 4 \cdot 5^x \). Differentiating with respect to \( x \): \[ f'(x) = 4 \cdot (5^x \log 5) \] Now, substitute \( x = -1 \): \[ f'(-1) = 4 \cdot (5^{-1} \log 5) \] \[ f'(-1) = 4 \cdot \frac{1}{5} \log 5 = \frac{4}{5} \log 5 \]
Step 4: Final Answer:
The value of \( f'(-1) \) is \( \frac{4}{5} \log 5 \).
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