Answer
$$L(x) \approx \frac{2 \sqrt{2}(\pi-4) }{\pi^{2}}x-\frac{\sqrt{2}\left(\frac{\pi}{2}-4\right)}{\pi}$$
Work Step by Step
Given
$$y=\frac{\sin x}{x}, \quad a=\frac{\pi}{4}$$
Since $y(\pi/4)=\dfrac{2 \sqrt{2}}{\pi}$ and \begin{align*} y'(x)& =\frac{x \cos x-\sin x}{x^{2}}\\ y'(\pi/4)&=\frac{2 \sqrt{2}(\pi-4)}{\pi^{2}} \end{align*} Then the linearization at $a=\pi/4$ is given by \begin{align*} L(x)&\approx y^{\prime}(a)(x-a)+y(a)\\ &= y^{\prime}(\pi/2)(x-\pi/4)+y(\pi/4)\\ &=\frac{2 \sqrt{2}(\pi-4)}{\pi^{2}}(x-\pi/4)+\frac{2 \sqrt{2}}{\pi}\\ &=\frac{2 \sqrt{2}(\pi-4) }{\pi^{2}}x-\frac{\sqrt{2}\left(\frac{\pi}{2}-4\right)}{\pi} \end{align*}
