Question:
The functions \(f, g\) and \(h\) satisfy the relations \( f'(x)=g(x+1) \) and \( g'(x)=h(x-1) \). Then \( f''(2x) \) is equal to
The functions \(f, g\) and \(h\) satisfy the relations \( f'(x)=g(x+1) \) and \( g'(x)=h(x-1) \). Then \( f''(2x) \) is equal to
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When derivatives are nested, repeatedly apply substitution carefully — especially shifting arguments like \(x+1\) or \(x-1\).
Updated On: May 8, 2026
- \(h(2x)\)
- \(4h(2x)\)
- \(h(2x-1)\)
- \(h(2x+1)\)
- \(2h'(2x)\)
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The Correct Option is A
Solution and Explanation
Concept:
We are given nested derivative relations:
\[
f'(x)=g(x+1), \quad g'(x)=h(x-1)
\]
To find \(f''(x)\), we differentiate \(f'(x)\), then substitute carefully using the given relations.
Step 1: Differentiate \(f'(x)\)
\[ f''(x) = \frac{d}{dx}[g(x+1)] \] Using chain rule: \[ f''(x) = g'(x+1) \]
Step 2: Use given relation for \(g'(x)\)
We know: \[ g'(x) = h(x-1) \] Replace \(x\) with \(x+1\): \[ g'(x+1) = h((x+1)-1) \]
Step 3: Simplify
\[ g'(x+1) = h(x) \] Thus: \[ f''(x) = h(x) \]
Step 4: Evaluate at \(2x\)
\[ f''(2x) = h(2x) \]
Step 5: Final Answer
\[ \boxed{h(2x)} \]
Step 1: Differentiate \(f'(x)\)
\[ f''(x) = \frac{d}{dx}[g(x+1)] \] Using chain rule: \[ f''(x) = g'(x+1) \]
Step 2: Use given relation for \(g'(x)\)
We know: \[ g'(x) = h(x-1) \] Replace \(x\) with \(x+1\): \[ g'(x+1) = h((x+1)-1) \]
Step 3: Simplify
\[ g'(x+1) = h(x) \] Thus: \[ f''(x) = h(x) \]
Step 4: Evaluate at \(2x\)
\[ f''(2x) = h(2x) \]
Step 5: Final Answer
\[ \boxed{h(2x)} \]
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