Question

Find dy/dx at x=0

Answer 1

\[\Large y=(1+x)(1+x^2)(1+x^4)..(1+x^2)^n\]

Answer 2

is that supposed to be 1+x^(2^n)

Answer 3

no,the question is correct

Answer 4

ok

Answer 5

take a logarithm on both sides

Answer 6

that gives 0

Answer 7

itll be easier

Answer 8

tried already

Answer 9

\[logy=\log(1+x)+\log(1+x^2)..\]

Answer 10

now
dy/ydx = d(...)/dx

Answer 11

\[\frac{dy}{dx}=y(\frac{1}{1+x}+\frac{1}{1+x^2}..)\]

Answer 12

the second term is 2x/(1=x^2)

Answer 13

oh yeah

Answer 14

so all otr terms will reduce to 0

Answer 15

so only y(1)/1+x

Answer 16

answer is 1 then?

Answer 17

what about "y"?

Answer 18

y is 1 at x=0

Answer 19

cool :D