f(x)=(x-1)e^2x I used the product rule: f'(x) = g'(x)h(x) + g(x)h'(x) = (1)(e^2x) + (x-1)(e^2x)(2) = e^2x + (2x-2)(e^2x) =e^2x [1 + (2x-2)] Where did I go wrong? The book answer is: f'(x) = e^2x(2x-1)
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Originally Posted by kmjt =e^2x [1 + (2x-2)] Where did I go wrong? The book answer is: f'(x) = e^2x(2x-1) You have the answer, the book just took an extra step $\displaystyle =e^{2x} [1 + (2x-2)]$ $\displaystyle =e^{2x} [1 +2x-2]$ $\displaystyle =e^{2x}[2x-1]$ $\displaystyle = e^{2x}(2x-1)$
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