I found \begin{align*} f'(x)&=-a\cos(\pi/2-ax-b) \\ f''(x)&= a^2 \cos(ax+b) \\ f'''(x)&=-a^3 \cos(\pi/2-ax-b) \\ f''''(x) &= a^4 \cos(ax+b) \\ f^{(n)} (x)&= (-1)^n a^n \cos(\dotsb +((-1)^n))(ax+b) \end{align*} I son t know what about the $\pi/2$ please help thanks
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You made a mistake at $f''(x)$, there's an extra minus. You then carried this mistake all the way. – user88595 Feb 07 '14 at 15:46
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1Hi, $f'(x)=-asin(ax+b)=acos(ax+b+\frac{\pi}2)$ – Feb 07 '14 at 15:46