# Did I differentiate this incorrectly?

• May 4th 2010, 08:15 PM
kmjt
Did I differentiate this incorrectly?
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)
• May 4th 2010, 08:30 PM
pickslides
Quote:

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

$=e^{2x} [1 + (2x-2)]$

$=e^{2x} [1 +2x-2]$

$=e^{2x}[2x-1]$

$= e^{2x}(2x-1)$