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Method of Differentiation: Let First Derivative

JEE Maths question with a full step-by-step solution.

Question
Let fn+1(x)=efn(x)f_{n+1}(x) = e^{f_n(x)} for all 1≤n≤91 \le n \le 9 and f1(x)=exf_1(x) = e^x. Then the first derivative of fn(x)f_n(x) for n=10n = 10 is:
Af9(x) f10(x)f_9(x)\, f_{10}(x)
B∏r=19fr(x) ex\prod_{r=1}^{9} f_r(x)\, e^x
C∏r=110exfr(x)\prod_{r=1}^{10} e^x f_r(x)
D∏r=110fr(x)\prod_{r=1}^{10} f_r(x)correct
Solution
Step 1: Differentiate the recurrence
fn+1′(x)=efn(x)fn′(x)=fn+1(x) fn′(x)f'_{n+1}(x) = e^{f_n(x)} f'_n(x) = f_{n+1}(x)\, f'_n(x)
 Step 2: Iterate the relation
f10′(x)=f10(x) f9′(x)=f10(x)f9(x) f8′(x)=⋯=f10(x)f9(x)⋯f2(x) f1′(x)f'_{10}(x) = f_{10}(x)\, f'_9(x) = f_{10}(x) f_9(x)\, f'_8(x) = \cdots = f_{10}(x) f_9(x) \cdots f_2(x)\, f'_1(x)
 Step 3: Use f1′(x)=ex=f1(x)f'_1(x) = e^x = f_1(x)
f10′(x)=f10(x)f9(x)⋯f2(x)f1(x)=∏r=110fr(x)f'_{10}(x) = f_{10}(x) f_9(x) \cdots f_2(x) f_1(x) = \prod_{r=1}^{10} f_r(x)
 Answer: (4)
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