Question

lf $f\left(x\right)={\tan }^{-1}x$, then $(1+x^2)f^{(n+2)}\left(x\right)+2x(n+1)f^{(n+1)}\left(x\right)+n(n+1)f^{\left(n\right)}\left(x\right)=$
$(n\ge 1,f^{\left(n\right)}(x)$ denotes the $n^{th}$ derivative of $f)$

• A. $0$
• B. $1$
• C. $f^n\left(x\right)+(1+x^2)f^{n+1}$
• D. $\left(n\right)(n+1)f^{n+1}\left(x\right)$

Answer 1

The correct option is A $0$

We have $f\left(x\right)={\tan }^{-1}x\Rightarrow f^1\left(x\right)=\frac{1}{1+x^2}\Rightarrow f^1\left(x\right)(1+x^2)=1$
Differentiating with respect to '$x$'
$f^2\left(x\right)(1+x^2)+f^1\left(x\right).2x=0$
Differentiating again with respect to '$x$',
$f^3\left(x\right)(1+x^2)+4x{f}^2\left(x\right)+2f^1\left(x\right)=0$ ......(1)
If we put $n=1$ in the given expression, we get it in the form of left hand side of equation (1).
Differentiating again with respect to '$x$', we get
$(1+x^2)f^4\left(x\right)+6x{f}^3\left(x\right)+6f^2\left(x\right)=0$.....(2)
So, for $n=2$ the expression is again equal to $0$.
Similarly, it can be proved by induction that the expression $(1+x^2)f^{(n+2)}\left(x\right)+2x(n+1)f^{(n+1)}\left(x\right)+n(n+1)f^{\left(n\right)}\left(x\right)=0$, for all $n\ge 1$