If `y=(1+x)(1+x^2)(1+x^4)(1+x^(2n)),` then find `(dy)/(dx)a tx=0.`

Options

A. 0
B. -1
C. 1
D. 2

Answer

C

Explanation

Given , ` y = (1 + x) (1 +x^(2)) (1 +x^(4)) …..(1 +x^(2n))`
On taking log both sides , we get
` log y = log[(1 + x)(1 + x^(2) )(1 + x^(4)) …(1 + x^(2n))]`
` rArr log y = log (1 +x)+ log(1 +x^(2)) + log (1 +x^(4)) + ...log(1 +x^(2n))`
On differentiating both sides w.r.t.x, we get
`(1)/(y) (dy)/(dx) = (1)/(1 +x) +(1)/(1 + x^(2)).2x + (1)/(1 +x^(4)).4x^(3) + ...+(1)/(1 +x^(2n)). 2nx^(2n-1)`
`rArr (dy)/(dx) = y [ (1)/(1+x)+(1)/(1 + x^(2)) +(4x^(3))/(1 + x^(4)) + ...+ (2nx^(2n-1))/(1 + x^(2n))]`
`rArr |(dy)/(dx)|_(x =0)=1[(1)/(1 +0) + 0 + 0 + ... + 0]`
`= 1, for x = 0 x , y= 1 `
Alternate Method
`y = ((1 -x) (1 +x)(1 +x^(2)) (1 +x^(4)) ......(1 + x^(2n)))/((1 -x))=(1 -x^(4n))/(1-x)`
` rArr (dy)/(dx) = ((1 -x) (-4nx^(4n-1))-(1-x^(4n))(-1))/((1-x)^(2))`
`= (-4n(1-x)x^(4n-1) + (1 -x^(4n)))/((1-x)^(2))`
` therefore ((dy)/*dx))_(x=0 =1`