Explanation The expression in Example 6 is y = 3 e 2 x x 2 + 1 . Replace x 2 + 1 with x + 1 in y = 3 e 2 x x 2 + 1 . That is, 3 e 2 x x + 1 . Consider g = 3 e 2 x x + 1 . Compute the derivative of g with respect to x as follows. d g d x = d d x ( 3 e 2 x x + 1 ) = 3 d d x ( e 2 x x + 1 ) [ ∵ d ( c ( g ( x ) ) ) d x = c d g d x ] = 3 ( ( ( x + 1 ) d ( e 2 x ) d x − e 2 x d ( x + 1 ) d x ) ( x + 1 ) 2 ) [ ∵ d d x ( u v ) = v d u d x − u d v d x v 2 ] = 3

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Chapter 27.6, Problem 2E

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Replace x2+1 with x + 1 in y=3e2xx2+1 and obtain the slope of a line tangent to the curve of g.

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Chapter 27.6, Problem 1E

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