Chapter 4 - Applications of the Derivative - 4.1 Linear Approximation and Applications - Exercises - Page 174: 53

$$L(x) \approx -\frac{4}{\pi^{2}} x+\frac{4}{\pi}$$

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