Find the fourth Maclaurin polynomial $P_4\left(x\right)$for the specified function:

$x\sin \left(x\right)$.

The function is,

$x\sin \left(x\right)$.

Let $f\left(x\right)=x\sin \left(x\right)$.

Since for any function $f$with a derivative of order $4$at $x=0$, the fourth Maclaurin polynomial is given by,

$P_4\left(x\right)=f\left(0\right)+f\text{'}\left(0\right)x+\frac{f\text{'}\text{'}\left(0\right)}{2!}{x}^2+\frac{f\text{'}\text{'}\text{'}\left(0\right)}{3!}{x}^3+\frac{f\text{'}\text{'}\text{'}\text{'}\left(0\right)}{4!}{x}^4$.

The value of the function at $x=0$is,

$$\begin{gathered} f\left(0\right)=0\sin \left(0\right) \\ =0 \end{gathered}$$

The derivatives of the function $f\left(x\right)=x\sin \left(x\right)$are,

$$\begin{gathered} f\text{'}\left(x\right)=\frac{d\left(x\sin \left(x\right)\right)}{dx} \\ =x\frac{d\left(\sin \left(x\right)\right)}{dx}+\sin \left(x\right)\frac{dx}{dx} \\ =x\cos \left(x\right)+\sin \left(x\right) \\ f\text{'}\left(0\right)=0\cos \left(0\right)+\sin \left(0\right) \\ =0+0 \\ =0 \end{gathered}$$

Also,

$$\begin{gathered} f\text{'}\text{'}\left(x\right)=\frac{d(x\cos \left(x\right)+\sin \left(x\right))}{dx} \\ =[x\frac{d\left(\cos \left(x\right)\right)}{dx}+\cos \left(x\right)\frac{dx}{dx}]+\frac{d\left(\sin \left(x\right)\right)}{dx} \\ =-x\sin \left(x\right)+\cos \left(x\right)+\cos \left(x\right) \\ =-x\sin \left(x\right)+2\cos \left(x\right) \\ f\text{'}\text{'}\left(0\right)=-0\sin \left(0\right)+2\cos \left(0\right) \\ =0+2 \\ =2 \end{gathered}$$

Also,

$$\begin{gathered} f\text{'}\text{'}\text{'}\left(x\right)=\frac{d(-x\sin \left(x\right)+2\cos \left(x\right))}{dx} \\ =-[x\frac{d\left(\sin \left(x\right)\right)}{dx}+\sin \left(x\right)\frac{dx}{dx}]+2\frac{d\left(\cos \left(x\right)\right)}{dx} \\ =-x\cos \left(x\right)-\sin \left(x\right)-2\sin \left(x\right) \\ =-x\cos \left(x\right)-3\sin \left(x\right) \\ f\text{'}\text{'}\text{'}\left(0\right)=-0\cos \left(0\right)-3\sin \left(0\right) \\ =-0-0 \\ =0 \end{gathered}$$

Also,

$$\begin{gathered} f\text{'}\text{'}\text{'}\text{'}\left(x\right)=\frac{d(-x\cos \left(x\right)-3\sin \left(x\right))}{dx} \\ =-[x\frac{d\left(\cos \left(x\right)\right)}{dx}+\cos \left(x\right)\frac{dx}{dx}]-3\frac{d\left(\sin \left(x\right)\right)}{dx} \\ =x\sin \left(x\right)-\cos \left(x\right)-3\cos \left(x\right) \\ =x\sin \left(x\right)-4\cos \left(x\right) \\ f\text{'}\text{'}\text{'}\text{'}\left(0\right)=0\sin \left(0\right)-4\cos \left(0\right) \\ =0-4 \\ =-4 \end{gathered}$$

The fourth Maclaurin polynomial is,

$$\begin{gathered} P_4\left(x\right)=0+0x+\frac{2}{2!}{x}^2+\frac{0}{3!}{x}^3+\frac{(-4)}{4!}{x}^4 \\ =x^2-\frac{1}{6}{x}^4 \end{gathered}$$