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**Unformatted text preview: **dy dx = d dx e x x + e x " # $ % & Differentiate. Use the quotient rule. dy dx = x + e x ( ) e x " e x 1 + e x ( ) x + e x ( ) 2 Simplify. dy dx = e x x + e x ( ) " 1 + e x ( ) [ ] x + e x ( ) 2 Factor. dy dx = e x x " 1 ( ) x + e x ( ) 2 Simplify the numerator. Now we evaluate the derivative at x = 0. dy dx x = = e x x " 1 ( ) x + e x ( ) 2 x = = e " 1 ( ) + e ( ) 2 = 1 " 1 ( ) + 1 ( ) 2 = " 1 1 = " 1 Now we know a point on the tangent line, (0, 1), and the slope of that line, -1. We will now use the point-slope form of a line to determine the equation of the desired tangent line. This is the point-slope form of a line. ( x 1 , y 1 ) = (0, 1) and m = -1. Simplify....

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