University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 3 - Section 3.5 - Derivatives of Trigonometric Functions - Exercises - Page 152: 63

Answer

The detailed explanations are below.
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Work Step by Step

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