Question:
If
\[
f(x)=
\begin{cases}
4(5^x), & x
8k+x, & x\geq 0
\end{cases}
\]
then \(f'(-1)=\)
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.
Updated On: May 7, 2026
- \(\frac{2}{5}\log 5\)
- \(\frac{4}{5}\log 5\)
- \(\frac{3}{5}\log 5\)
- \(20\log 5\)
Show Solution
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The Correct Option is B
Solution and Explanation
Concept:
For a piecewise function, choose the correct branch according to the point at which derivative is required.
Step 1: We need: \[ f'(-1) \] Since: \[ -1<0 \] we will use the first branch: \[ f(x)=4(5^x) \]
Step 2: Differentiate \(f(x)=4(5^x)\). \[ \frac{d}{dx}(5^x)=5^x\log 5 \] Therefore: \[ f'(x)=4\cdot 5^x\log 5 \]
Step 3: Put \(x=-1\). \[ f'(-1)=4\cdot 5^{-1}\log 5 \] \[ f'(-1)=4\cdot \frac{1}{5}\log 5 \] \[ f'(-1)=\frac{4}{5}\log 5 \] Therefore, \[ \boxed{\frac{4}{5}\log 5} \]
Step 1: We need: \[ f'(-1) \] Since: \[ -1<0 \] we will use the first branch: \[ f(x)=4(5^x) \]
Step 2: Differentiate \(f(x)=4(5^x)\). \[ \frac{d}{dx}(5^x)=5^x\log 5 \] Therefore: \[ f'(x)=4\cdot 5^x\log 5 \]
Step 3: Put \(x=-1\). \[ f'(-1)=4\cdot 5^{-1}\log 5 \] \[ f'(-1)=4\cdot \frac{1}{5}\log 5 \] \[ f'(-1)=\frac{4}{5}\log 5 \] Therefore, \[ \boxed{\frac{4}{5}\log 5} \]
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