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lecture21bc

# lecture21bc - dy dx = d dx e x x e x" \$& ’...

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§ 4.2 The Exponential Function e x Math 121 Lecture 21 ! e ! The Derivatives of 2 x , b x , and e x

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Definition Example e : An irrational number, approximately equal to 2.718281828, such that the function f ( x ) = b x has a slope of 1, at x = 0, when b = e
Calculate.

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Differentiate y = ( x + e x ) 4 This is the given function. dy dx = 4 " ( x + e x ) 3 d dx x + e x ( ) Differentiate (generalized power rule). = 4( x + e x ) 3 (1 + e x ) Since y = ( x + e x ) 4 d dx e x = e x Differentiate y = ( x + 1) 2 e x This is the given function. dy dx = ( x + 1) 2 d dx e x ( ) + e x d dx ( x + 1) 2 Differentiate (product rule). = ( x + 1) 2 e x + e x " 2 " ( x + 1) " d dx ( x + 1) = (( x + 1) 2 + 2 x + 2) e x = ( x 2 + 4 x + 3) e x Since d dx e x = e x y = ( x + 1) 2 e x
Find the equation of the tangent line to the curve at (0, 1). We must first find the derivative function and then find the value of the derivative at (0, 1). Then we can use the point-slope form of a line to find the desired tangent line equation. This is the given function.

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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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