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Chapter 3 | Derivatives
By applying the limit definition of the derivative to j ( x ) = f ( x ) g ( x ), we obtain j ′( x ) = lim h →0 f ( x + h ) g ( x + h )− f ( x ) g ( x ) h . By adding and subtracting f ( x ) g ( x + h ) in the numerator, we have j ′( x ) = lim h →0 f ( x + h ) g ( x + h )− f ( x ) g ( x + h )+ f ( x ) g ( x + h )− f ( x ) g ( x ) h . After breaking apart this quotient and applying the sum law for limits, the derivative becomes j ′( x ) = lim h →0 ⎛ ⎝ f ( x + h ) g ( x + h )− f ( x ) g ( x + h ) h ⎞ ⎠ + lim h →0 ⎛ ⎝ f ( x ) g ( x + h )− f ( x ) g ( x ) h ⎞ ⎠ . Rearranging, we obtain j ′( x ) = lim h →0 ⎛ ⎝ f ( x + h )− f ( x ) h · g ( x + h ) ⎞ ⎠ + lim h →0 ⎛ ⎝ g ( x + h )− g ( x ) h · f ( x ) ⎞ ⎠ . By using the continuity of g ( x ), the definition of the derivatives of f ( x ) and g ( x ), and applying the limit laws, we arrive at the product rule, j ′( x ) = f ′( x ) g ( x )+ g ′( x ) f ( x ). □ Example 3.23 Applying the Product Rule to Functions at a Point
For j ( x ) = f ( x ) g ( x ), use the product rule to find j ′(2) if f (2) =3, f ′(2) =−4, g (2) =1, and g ′(2) =6.
Solution Since j ( x ) = f ( x ) g ( x ), j ′( x ) = f ′( x ) g ( x )+ g ′( x ) f ( x ), and hence
j ′(2) = f ′(2) g (2)+ g ′(2) f (2) = (−4)(1) + (6)(3) = 14.
Example 3.24 Applying the Product Rule to Binomials
For j ( x ) = ( x 2 +2)(3 x 3 −5 x ), find j ′( x ) by applying the product rule. Check the result by first finding the product and then differentiating. Solution If we set f ( x ) = x 2 +2 and g ( x ) =3 x 3 −5 x , then f ′( x ) =2 x and g ′( x ) =9 x 2 −5. Thus, j ′( x ) = f ′( x ) g ( x )+ g ′( x ) f ( x ) = (2 x ) ⎛ ⎝ 3 x 3 −5 x ⎞ ⎠ +(9 x 2 −5)( x 2 +2). Simplifying, we have j ′( x ) =15 x 4 +3 x 2 −10.
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