find the linearization of the function $k(x)=(x^{2}+1)^{-2}$ at $x = 1$.

Explanation:

Step1: Find the derivative

Let $u = x^{2}+1$, then $k(x)=u^{-2}$. Using chain - rule, $k^\prime(x)=-2u^{-3}\cdot2x=-4x(x^{2}+1)^{-3}$.

Step2: Evaluate $k(1)$ and $k^\prime(1)$

$k(1)=(1^{2}+1)^{-2}=\frac{1}{4}$, $k^\prime(1)=-4\times1\times(1^{2}+1)^{-3}=-\frac{1}{2}$.

Step3: Use linearization formula

The linearization $L(x)=k(1)+k^\prime(1)(x - 1)$. So $L(x)=\frac{1}{4}-\frac{1}{2}(x - 1)$.

Answer:

$L(x)=\frac{1}{4}-\frac{1}{2}(x - 1)$