Chapter 10: Infinite Sequences and Series - Section 10.8 - Taylor and Maclaurin Series - Exercises 10.8 - Page 619: 43

$L(x)=1$ and $Q(x)=1+\dfrac{x^2}{2!}$

Differentiate the given function $f(x)$ as follows: $f'(x)=\dfrac{x}{(1-x^2)^{3/2}} ; \\f''(x)=\dfrac{1}{(1-x^2)^{3/2}}+\dfrac{3x^2}{(1-x^2)^{5/2}}$ Now, $ f(0)=1; \\ f'(0)=1; \\ f''(0)=1$ Therefore, $L(x)=f(0)+xf'(0)=1; \\ Q(x) =f(0) x +xf'(0) +\dfrac{x^2 f''(0)}{2!}=1+\dfrac{x^2}{2!}$