Explanation Given: The function is g ( x ) = ( 2 x 2 + 3 ) ( 5 x + 2 ) 6 x − 7 . Formula used: (1) The quotient rule is: d d x [ u ( x ) v ( x ) ] = v ( x ) ⋅ u ' ( x ) − u ( x ) ⋅ v ' ( x ) ( v ( x ) ) 2 . (2). The product rule is: d d x [ u ( x ) ⋅ v ( x ) ] = u ( x ) ⋅ v ' ( x ) + v ( x ) ⋅ u ' ( x ) . (3). The sum rule : d d x ( u ( x ) + v ( x ) ) = d d x ( u ( x ) ) + d d x ( v ( x ) ) . (4). The difference rule : d d x ( u ( x ) + v ( x ) ) = d d x ( u ( x ) ) + d d x ( v ( x ) ) . (5). The power rule: d d x ( x n ) = n x n − 1 . (6). The constant rule: d d x ( k ) = 0 . Calculation: Let u ( x ) = ( 2 x 2 + 3 ) ( 5 x + 2 ) and v ( x ) = 6 x − 7 . Apply the quotient rule for the given function g ( x ) = ( 2 x 2 + 3 ) ( 5 x + 2 ) 6 x − 7 as follows. g ' ( x ) = d d x [ ( 2 x 2 + 3 ) ( 5 x + 2 ) 6 x − 7 ] = ( 6 x − 7 ) d d x ( ( 2 x 2 + 3 ) ( 5 x + 2 ) ) − ( ( 2 x 2 + 3 ) ( 5 x + 2 ) ) d d x ( 6 x − 7 ) ( 6 x − 7 ) 2 Apply the product rule for the term d d x ( ( 2 x 2 + 3 ) ( 5 x + 2 ) ) as follows. g ' ( x ) = ( 6 x − 7 ) ( ( 5 x + 2 ) d d x ( 2 x 2 + 3 ) + ( 2 x 2 + 3 ) d d x ( ( 5 x + 2 ) ) ) ( 6 x − 7 ) 2 − ( ( 2 x 2 + 3 ) ( 5 x + 2 ) ) d d x ( 6 x − 7 ) ( 6 x − 7 ) 2 Apply the sum and difference rule as follows

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Chapter 12.2, Problem 28E

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To find: The derivative of the function g(x)=(2x2+3)(5x+2)6x−7 .

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Chapter 12.2, Problem 27E

Chapter 12.2, Problem 29E

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For the following function f and real number a, a. find the slope of the tangent line mtan = f' (a), and b. find the equation of the tangent line to f at x = a. f(x)= 2 = a = 2 x2 a. Slope: b. Equation of tangent line: y

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The derivative of the function g ( x ) = ( 2 x 2 + 3 ) ( 5 x + 2 ) 6 x − 7 .

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To find: The derivative of the function g(x)=(2x2+3)(5x+2)6x−7 .

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Chapter 12 Solutions