## University Calculus: Early Transcendentals (3rd Edition)

We will prove that the slopes of these two tangent lines multiply to $-1$.
$$y=\frac{\pi \sin x}{x}$$ 1) First, find the derivative of $y$: $$y'=\frac{(\pi \sin x)'x-(\pi\sin x)x'}{x^2}$$ $$y'=\frac{\pi x\cos x-\pi \sin x}{x^2}$$ 2) For $x=\pi$, we have: $$y'(\pi)=\frac{\pi\times\pi\cos\pi-\pi\sin\pi}{\pi^2}=\frac{\pi^2(-1)-\pi\times0}{\pi^2}$$ $$=\frac{-\pi^2}{\pi^2}=-1$$ For $x=-\pi$, we have: $$y'(-\pi)=\frac{\pi\times(-\pi)\cos(-\pi)-\pi\sin(-\pi)}{\pi^2}=\frac{-\pi^2(-1)-\pi\times0}{\pi^2}$$ $$=\frac{\pi^2}{\pi^2}=1$$ Since the product of the two slopes: $(-1)\times1=-1$, the tangent lines at $x=\pi$ and $x=-\pi$ must be perpendicular to each other, or we can say, intersect at right angles.