Chapter 3 | Derivatives
251
derivative of f and the derivative of g : d dx ⎛
⎠ = d
⎞ ⎠ − d
⎝ f ( x )− g ( x ) ⎞
⎛ ⎝ f ( x )
⎛ ⎝ g ( x )
⎞ ⎠ ;
dx
dx
that is,
for j ( x ) = f ( x )− g ( x ), j ′( x ) = f ′( x )− g ′( x ). Constant Multiple Rule . The derivative of a constant k multiplied by a function f is the same as the constant multiplied by the derivative: d dx ⎛ ⎝ kf ( x ) ⎞ ⎠ = k d dx ⎛ ⎝ f ( x ) ⎞ ⎠ ; that is, for j ( x ) = kf ( x ), j ′( x ) = kf ′( x ). Proof We provide only the proof of the sum rule here. The rest follow in a similar manner. For differentiable functions f ( x ) and g ( x ), we set j ( x ) = f ( x )+ g ( x ). Using the limit definition of the derivative we have j ′( x ) = lim h →0 j ( x + h )− j ( x ) h . By substituting j ( x + h ) = f ( x + h )+ g ( x + h ) and j ( x ) = f ( x )+ g ( x ), we obtain j ′( x ) = lim h →0 ⎛ ⎝ f ( x + h )+ g ( x + h ) ⎞ ⎠ − ⎛ ⎝ f ( x )+ g ( x ) ⎞ ⎠ h .
Rearranging and regrouping the terms, we have j ′( x ) = lim h →0 ⎛ g ( x + h )− g ( x ) h ⎞ ⎠ . We now apply the sum law for limits and the definition of the derivative to obtain j ′( x ) = lim h →0 ⎛ ⎝ f ( x + h )− f ( x ) h ⎞ ⎠ + lim h →0 ⎛ ⎝ g ( x + h )− g ( x ) h ⎞ ⎝ f ( x + h )− f ( x ) h +
⎠ = f ′( x )+ g ′( x ).
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Example 3.20 Applying the Constant Multiple Rule
Find the derivative of g ( x ) =3 x 2 and compare it to the derivative of f ( x ) = x 2 .
Solution We use the power rule directly:
g ′( x ) = d dx ⎛ ⎞ ⎠ =3(2 x ) =6 x . Since f ( x ) = x 2 has derivative f ′( x ) =2 x , we see that the derivative of g ( x ) is 3 times the derivative of ⎝ 3 x 2 ⎞ ⎠ =3 d dx ⎛ ⎝ x 2
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