Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 3 - Differentiation - 3.6 Trigonometric Functions - Exercises - Page 140: 43

Answer

$f'(x)=-\sin{x}$ $f''(x)=-\cos{x}$ $f'''(x)=\sin{x}$ $f^{(4)}(x)=\cos{x}$ $f^{(5)}(x)=-\sin{x}$ $f^{(8)}(x)=\cos{x}$ $f^{(37)}(x)=-\sin{x}$

Work Step by Step

Firstly find the first derivative of $f (x) = cos x$. That is, $f'(x)=(\cos{x})'=-\sin{x}$ or $f'(x)=-1\times \sin{x}$ Now again differentiate using the product rule to find the second derivative. $f''(x)=(-1\times \sin{x})'$ $\hspace{45px}=(-1)'\sin{x}+(\sin{x})'(-1)$ $\hspace{45px}=0\times\sin{x}+\cos{x}(-1)$ $\hspace{45px}=-\cos{x}$ Now again differentiate using the product rule to find the third derivative. $f'''(x)=(-1\times \cos{x})'$ $\hspace{45px}=(-1)'\cos{x}+(\cos{x})'(-1)$ $\hspace{45px}=0\times\cos{x}+(-\sin{x})(-1)$ $\hspace{45px}=\sin{x}$ Now again differentiate to find the fourth derivative. $f^{(4)}(x)=( \sin{x})'=\cos{x}$ Now again differentiate to find the fifth derivative. $f^{(5)}(x)=( \cos{x})'=-\sin{x}$ We see that after derivating four times, we get the same value. That is, $f^{(n+4)}(x)=f^{(n)}(x)$ Now, substitute $n=4$. $f^{(8)}(x)=f^{(4)}(x)$ Now, substitute $f^{(4)}(x)=\cos{x}$. We get, $f^{(8)}(x)=\cos{x}$. Similarly, $f^{(37)}(x)=f^{(33)}(x)=f^{(29)}(x)=f^{(25)}(x)=f^{(21)}(x)=f^{(17)}(x)=f^{(13)}(x)=f^{(9)}(x)=f^{(5)}(x)$ Hence, we get $f^{(37)}(x)=f^{(5)}(x)$. Now, substitute $f^{(5)}(x)=-\sin{x}$. We get, $f^{(37)}(x)=-\sin{x}$.
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