Differentiate it again and again (it's fun)

$\frac { dy }{ dx } =\frac { 1 }{ 1+x^{ 2 } } =1-{ x }^{ 2 }+{ x }^{ 4 }-{ x }^{ 6 }+.......+{ x }^{ 20 }-.................$

$\frac { { d }^{ 21 }y }{ { dx }^{ 21 } } =0+20!+term\quad consisting\quad x\quad ....$

${ \left( \frac { { d }^{ 21 }y }{ { dx }^{ 21 } } \right) }_{ x=0 }=\quad 20!$

So $k=\fbox{20}$

sir i think u r asking 21st derivative at x=0...

my answer was using taylor series of arctanx which makes derivatives easy.

$\arctan x = x - \frac{x^{3}}{3} + \frac{x^{5}}{5} + \dots -\frac{x^{19}}{19} + \frac{x^{21}}{21} - \frac{x^{23}}{ 23} + \dots$

Now differentiate w.r.t $x$ 21 times

$Ans=\frac{21!}{21} - \frac{23! x^2 }{23\cdot 2!} + \dots \quad \text{terms with x}$

Now put $x=0$ and u get the answer as $\frac{21!}{21}=20!$

(1+x^2)y#1=1 after first differentiation. (Here, the number after the hash indicates derivative). Using Leibnitz theorem for repeated derivatives, at x=0 we can develop the recurrence formula y#(n+1)= -n(n-1)y#(n-1). Hence the value of y#21