Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 2 - Differentiation - 2.4 Exercises - Page 139: 113

Answer

(a) $f' = \beta \cos(\beta x) , f'' = - \beta^{2} \sin(\beta x), f''' = -\beta^{3}\cos(\beta x) , f'''' = \beta^{4}\sin(\beta x) $ (b) $f'' + \beta^{2}f = - \beta^{2} \sin(\beta x) + \beta^{2} \sin(\beta x) = 0 $ so the equation is satisfied (c) $f^{(2k)} = (-1)^{k} * \beta^{2k}*\sin(\beta x), f^{(2k-1)} = (-1)^{k-1}*\beta^{2k-1}*\cos(\beta x)$

Work Step by Step

(a) Since $f = \sin(\beta x)$, we apply the chain rule to find $f'$ in the following manner (knowing that the derivative of cosine is negative sine and the derivative of sine is cosine): $f' = \frac{d(\beta x)}{dx} *\frac{d(\sin(\beta x))}{d(\beta x)} =\frac{d(\beta x)}{dx} * (\cos(\beta x)) = \beta * \cos(\beta x) $ Similarly for the higher derivatives, we apply the chain rule repeatedly: $f'' = \beta * \frac{d(\beta x)}{dx} * (- \sin(\beta x)) = - \beta^{2} \sin(\beta x)$ $f''' = \beta^{2} * \frac{d(\beta x)}{dx} * (- ( \cos(\beta x)) = -\beta^{3}\cos(\beta x) $ $ f'''' = \beta^{3} * \frac{d(\beta x)}{dx} * (- ( -\sin(\beta x)) = \beta^{4}\sin(\beta x) $ (b) As shown above, when we substitute in $f$ and $f''$ : $f'' + \beta^{2}f = - \beta^{2} \sin(\beta x) + \beta^{2} \sin(\beta x) = 0 $ so the equation is satisfied (c) As can be seen above, since the derivative of cosine is negative sine and the derivative of sine is cosine, the even derivatives $f^{(2k)}$ alternate in sign, starting with $f''$ being negative. Also, since each derivative brings out a single power of $\beta$ due to the chain rule, $f^{(2k)}$ must have a multiple of $\beta^{(2k)}$. Finally, because all of the even derivatives result in some multiple of $\sin(\beta x)$ (since the derivative of cosine is negative sine and the derivative of sine is cosine), we may combine all of this information to find: $f^{(2k)} = (-1)^{k} * \beta^{2k}*\sin(\beta x)$ Similarly, the odd derivatives $f^{(2k -1)}$ alternate in sign, starting with $f'$ being positive. Also, since each derivative brings out a single power of $\beta$ due to the chain rule, $f^{(2k-1)}$ must have a multiple of $\beta^{(2k-1)}$. Finally, because all of the odd derivatives result in some multiple of $\cos(\beta x)$ (since the derivative of cosine is negative sine and the derivative of sine is cosine), we may combine all of this information to find: $f^{(2k-1)} = (-1)^{k-1}*\beta^{2k-1}*\cos(\beta x)$
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