## Calculus (3rd Edition)

$y=\frac{4\sqrt 2}{\pi}x-\sqrt 2$
The equation of the tangent line is of the form $y=f(a)+f'(a)(x-a)$ Since $a=\frac{\pi}{4}$, $f(a)=f(\frac{\pi}{4})=\frac{\sin\frac{\pi}{4}-\cos\frac{\pi}{4}}{\frac{\pi}{4}}=0$ $f'(\theta)=\frac{(\cos\theta+\sin\theta)\theta-(\sin\theta-\cos\theta)1}{\theta^{2}}$ (use quotient rule) $f'(a)=f'(\frac{\pi}{4})=\frac{(\frac{1}{\sqrt 2}+\frac{1}{\sqrt 2})\frac{\pi}{4}-0}{(\frac{\pi}{4})^{2}}=\frac{4\sqrt 2}{\pi}$ Therefore, the equation of the tangent line is $y=0+\frac{4\sqrt 2}{\pi}(x-\frac{\pi}{4})$ Or $y=\frac{4\sqrt 2}{\pi}x-\sqrt 2$