Chapter 9 - Section 9.8 - Taylor and Maclaurin Series - Exercises - Page 537: 42

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

We have $f(x) = e^{\sin x}$ and $f'(x)=-\cos x e^{\sin x} ; \\f''(x)=-(\sin x) e^{\sin x}+(\cos x)^2 e^{\sin x} $ $\implies f(0)=1; f'(0)=1; f''(0)=1$ Therefore, $L(x)=f(0)+xf'(0)=1+x$ and $Q(x) =f(0) x +xf'(0) +\dfrac{x^2}{2!}f''(0)=1+x+\dfrac{x^2}{2!}$ So, $L(x)=1+x$ and $Q(x)=1+x+\dfrac{x^2}{2!}$