Question:medium

If \[ f(x)= \begin{cases} 4(5^x), & x 8k+x, & x\geq 0 \end{cases} \] then \(f'(-1)=\)

Show Hint

For piecewise functions, first check which interval contains the given value of \(x\), then differentiate that branch only.
  • \(\frac{2}{5}\log 5\)
  • \(\frac{4}{5}\log 5\)
  • \(\frac{3}{5}\log 5\)
  • \(20\log 5\)
Show Solution

The Correct Option is B

Solution and Explanation

Step 1: Understanding the Question:
This problem involves a piecewise function. To find the derivative at a specific point, we must first identify which "piece" or interval the point falls into.
The point in question is \( x = -1 \). Since \( -1<0 \), we only need to consider the first part of the function definition, which is \( f(x) = 4(5^x) \).
Step 2: Key Formula or Approach:
1. Derivative of exponential function \( a^x \): \( \frac{d}{dx}(a^x) = a^x \ln a \).
2. Power rule for constants: \( \frac{d}{dx}(c \cdot g(x)) = c \cdot g'(x) \).
Step 3: Detailed Explanation:

For \( x<0 \), the function is defined as \( f(x) = 4 \cdot 5^x \).

We find the general derivative expression for this piece:
\[ f'(x) = \frac{d}{dx}(4 \cdot 5^x) = 4 \cdot \frac{d}{dx}(5^x) \]

Applying the exponential derivative rule with base 5:
\[ f'(x) = 4 \cdot 5^x \cdot \ln 5 \]

Now, we substitute the value \( x = -1 \) into this derivative formula:
\[ f'(-1) = 4 \cdot 5^{-1} \cdot \ln 5 \]

Simplify the term \( 5^{-1} \):
\[ f'(-1) = 4 \cdot \frac{1}{5} \cdot \ln 5 \]
\[ f'(-1) = \frac{4}{5} \ln 5 \]

Note: In these options, "log" refers to the natural logarithm \( \ln \).

Step 4: Final Answer:
The value of \( f'(-1) \) is \( \frac{4}{5} \log 5 \).
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