\text { If } y=(1+x)\left(1+x^{2}\right)\left(1+x^{4}\right) \ldots

Question

\text { If } y=(1+x)\left(1+x^{2}\right)\left(1+x^{4}\right) \ldots

\text { If } y=(1+x)\left(1+x^{2}\right)\left(1+x^{4}\right) \ldots .\left(1+x^{2 n}\right) \text { then the value of }\left(\frac{d y}{d x}\right) \text { at } x=0 \text { is }\(\text { If } y=(1+x)\left(1+x^{2}\right)\left(1+x^{4}\right) \ldots .\left(1+x^{2 n}\right) \text { then the value of }\left(\frac{d y}{d x}\right) \text { at } x=0 \text { is }\)

Monjur Alahi

4 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Salim Ahmed Kawsar

y=(1+x)(1+x²)(1+x⁴)......(1+x²ⁿ)

⇒log y =Iog (1+x)+ Iog (1+x²) ......+ Iog (1+x²ⁿ)
⇒1/y. dy/dx= 1/(1+x )+ 2x/(1+x²)......+ 2nx²ⁿ⁻¹/(1+x²ⁿ)

⇒dy/dx= y.(1/1+x + 2x/1+x².......+ 2nx²ⁿ⁻¹/1+x²ⁿ)
Substitute the value x=0,
⇒∣dy/dx∣_x=0 =1.1 = 1

4 views Answered 2023-02-05 15:29

\text { If } y=(1+x)\left(1+x^{2}\right)\left(1+x^{4}\right) \ldots

1 Answers

y=(1+x)(1+x2)(1+x4)......(1+x2n) ⇒log y =Iog (1+x)+ Iog (1+x2) ......+ Iog (1+x2n) ⇒1/y. dy/dx= 1/(1+x )+ 2x/(1+x2 )......+ 2nx2n−1/(1+x2n) ⇒dy/dx= y.(1/1+x + 2x/1+x2.......+ 2nx2n−1/1+x2n) Substitute the value x=0, ⇒∣dy/dx∣x=0 =1.1 = 1

Related Questions

y=(1+x)(1+x²)(1+x⁴)......(1+x²ⁿ)

⇒log y =Iog (1+x)+ Iog (1+x²) ......+ Iog (1+x²ⁿ)
⇒1/y. dy/dx= 1/(1+x )+ 2x/(1+x²)......+ 2nx²ⁿ⁻¹/(1+x²ⁿ)

⇒dy/dx= y.(1/1+x + 2x/1+x².......+ 2nx²ⁿ⁻¹/1+x²ⁿ)
Substitute the value x=0,
⇒∣dy/dx∣_x=0 =1.1 = 1