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

The graphs of $y=-\sin x$ and $y=\frac{\cos(x+h)-\cos x}{h}$ are below. 1) On the left graph, as $h$ gets larger from $-1$ (the blue curve) to $-0.3$ (the purple curve), the curve of the function $y=\frac{\cos(x+h)-\cos x}{h}$ gets closer and apparently matches up with the curve of the function $y=-\sin x$. We can generalize this by saying that as $h\to0^-$, $$\lim_{h\to0^-}\frac{\cos(x+h)-\cos x}{h}=-\sin x$$ 2) Similarly, on the right graph, as $h$ gets smaller from $1$ (the blue curve) to $0.1$ (the black curve), the curve of the function $y=\frac{\cos(x+h)-\cos x}{h}$ gets closer and apparently matches up with the curve of the function $y=-\sin x$. We can generalize this by saying that as $h\to0^+$, $$\lim_{h\to0^+}\frac{\cos(x+h)-\cos x}{h}=-\sin x$$ Therefore, as the left-hand and right-hand limits are equal, $$\lim_{h\to0}\frac{\cos(x+h)-\cos x}{h}=-\sin x$$ However, according to the definition of derivative, we also have $$\lim_{h\to0^+}\frac{\cos(x+h)-\cos x}{h}=\frac{d}{dx}(\cos x)$$ That means, $$\frac{d}{dx}(\cos x)=-\sin x$$ So this exercise demonstrates the fact that the derivative of $\cos x$ is $-\sin x$.