If y=tan−1⁡11+x+x2+tan−1⁡1x2+3x+3+tan−1⁡1x2+5x+7+…. upto n terms then find  the value of y1(0)

y=Tan−1⁡x+1−x1+x(1+x)+Tan−1⁡(x+2)−(x+1)1+(x+1)(x+2)+…n terms =Tan−1⁡(x+1)−Tan−1⁡x+Tan−1⁡(x+2)−Tan−1⁡(x+1_)+….++Tan−1⁡(x+n)−Tan−1⁡(x+(n−1))y=Tan−1⁡(x+n)−Tan−1⁡xy′=11+(x+n)2−11+x2⇒y′(0)=11+n2−1=1−1−n21+n2y′(0)=−n21+n2