University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 3 - Section 3.6 - The Chain Rule - Exercises - Page 160: 106

Answer

As $h\to0$, the curve $y=\frac{\cos((x+h)^2)-\cos(x^2)}{h}$ gets closer and closer to the curve $y=-2x\sin(x^2)$, apparently approaching to become one with $y=-2x\sin(x^2)$.
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Work Step by Step

The graphs of the function are enclosed below. The graph of $y=-2x\sin(x^2)$ is portrayed by the black curve, while the other curves are variations of $y=\frac{\cos((x+h)^2)-\cos(x^2)}{h}$. From the graph, we see that as $h\to0$, the curve $y=\frac{\cos((x+h)^2)-\cos(x^2)}{h}$ gets closer and closer to the curve $y=-2x\sin(x^2)$, apparently approaching to become one with $y=-2x\sin(x^2)$. In other words, we can say that $$\lim_{h\to0}\frac{\cos((x+h)^2)-\cos(x^2)}{h}=-2x\sin(x^2)$$ From the definition of derivative, we know that $$\lim_{h\to0}\frac{\cos((x+h)^2)-\cos(x^2)}{h}=(\cos (x^2))'$$ Therefore, $$(\cos(x^2))'=-2x\sin(x^2)$$ So the behavior of the graphs can be summarized to show that the derivative of $\cos(x^2)$ is $-2x\sin(x^2)$.
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