If `y=tan^(-1)1/(1+x+x^2)+tan^(-1)1/(x^2+3x+3)+tan^(-1)1/(x^2+5x+7)++` upto `n` terms, then find the value of `y^(prime)(0)dot`

Options

A. `-(1)/(1 + n^(2))`
B. `- (n^(2))/(1 + n^(2))`
C. `(n)/(1 + n^(2))`
D. Nonw of these

Answer

B

Explanation

`y = sum_(r=1)^(n) tan^(-1) {(1)/(1+ (x + r-1)(x +r))} `
` = sum_(r=1)^(n) tan^(-1) {((x + r)- (x +r -1))/(1+ (x + r-1)(x +r))} `
` = sum_(r=1)^(n) tan^(-1) {tan^(-1) (x + r) - tan^(-1) (x +r -1)}`
` = tan^(-1) (x + n) - tan^(-1) x`
`y'=(1)/(1 + (x + n)^(2)) - (1)/(1 + x^(2))`
` therefore y(0) = (1)/(1 + n^(2)) - 1 = (-n^(2))/(1 + n^(2))`