Here's my answer... f'(x) = [(g'(x)) . (h(x)) - g(x) . h'(x)]/ (h(x))^2 = [(e^x - e^-x) (2) - (e^x + e^-x) (0)]/4 = 2(e^x - e^-x)/4 = (e^x - e^-x)/2 f'(0) = [e^0 - e^0]/2 = 0 Did I do that right?
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Originally Posted by satx Here's my answer... f'(x) = [(g'(x)) . (h(x)) - g(x) . h'(x)]/ (h(x))^2 = [(e^x - e^-x) (2) - (e^x + e^-x) (0)]/4 = 2(e^x - e^-x)/4 = (e^x - e^-x)/2 f'(0) = [e^0 - e^0]/2 = 0 Did I do that right? i believe you did it right
Originally Posted by satx Here's my answer... f'(x) = [(g'(x)) . (h(x)) - g(x) . h'(x)]/ (h(x))^2 = [(e^x - e^-x) (2) - (e^x + e^-x) (0)]/4 = 2(e^x - e^-x)/4 = (e^x - e^-x)/2 f'(0) = [e^0 - e^0]/2 = 0 Did I do that right? Very inefficient and plenty of opportunity to make careless mistakes. Since $\displaystyle f(x) = \frac{e^x}{2} + \frac{e^{-x}}{2}$, why not just differentiate term-by-term?
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