Method of DifferentiationmediumFree

Method of Differentiation — JEE Maths practice question

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

Question
If f(x)=(1+x)(1+x2)(1+x4)⋯(1+x2n)f(x)=(1+x)(1+x^{2})(1+x^{4})\cdots(1+x^{2^{n}}), then at x=0x=0, f′(x)f'(x) is
A11correct
B−1-1
C00
D22
Solution
Step 1: Multiply both sides by (1−x)(1-x):
(1−x)f(x)=(1−x)(1+x)(1+x2)(1+x4)⋯(1+x2n)(1-x)f(x)=(1-x)(1+x)(1+x^{2})(1+x^{4})\cdots(1+x^{2^{n}})
 Using the telescoping product (1−x)(1+x)=1−x2(1-x)(1+x)=1-x^{2}, (1−x2)(1+x2)=1−x4(1-x^{2})(1+x^{2})=1-x^{4}, and so on:
(1−x)f(x)=1−x2n+1(1-x)f(x)=1-x^{2^{n+1}}
f(x)=1−x2n+11−xf(x)=\dfrac{1-x^{2^{n+1}}}{1-x}
 Step 2: Differentiate using the quotient rule:
f′(x)=−2n+1x2n+1−1(1−x)+(1−x2n+1)(1−x)2f'(x)=\dfrac{-2^{n+1}x^{2^{n+1}-1}(1-x)+(1-x^{2^{n+1}})}{(1-x)^{2}}
 Step 3: At x=0x=0: x2n+1−1=0x^{2^{n+1}-1}=0 and x2n+1=0x^{2^{n+1}}=0:
f′(0)=0+112=1f'(0)=\dfrac{0+1}{1^{2}}=1
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