Calculus Volume 1

Chapter 3 | Derivatives

233

Substitute f ( x + h ) = x + h and f ( x ) = x into f ′( x ) = lim h →0 f ( x + h )− f ( x ) h . Multiply numerator and denominator by x + h + x without distributing in the denominator. Multiply the numerators and simplify.

x + h − x h

f ′( x ) = lim h →0

x + h − x h

x + h + x x + h + x

= lim

·

h →0

h h ( x + h + x ) 1 ( x + h + x )

= lim

h →0

= lim

Cancel the h .

h →0

= 1 2 x

Evaluate the limit.

Example 3.12 Finding the Derivative of a Quadratic Function

Find the derivative of the function f ( x ) = x 2 −2 x .

Solution Follow the same procedure here, but without having to multiply by the conjugate.

Substitute f ( x + h ) = ( x + h ) 2 −2( x + h ) and f ( x ) = x 2 −2 x into f ′( x ) = lim h →0 f ( x + h )− f ( x ) h .

(( x + h ) 2 −2( x + h ))−( x 2 −2 x ) h

f ′( x ) = lim h →0

x 2 +2 xh + h 2 −2 x −2 h − x 2 +2 x h

x + h ) 2 −2( x + h ).

= lim

Expand(

h →0

2 xh −2 h + h 2 h h (2 x −2+ h ) h (2 x −2+ h )

= lim

Simplify.

h →0

= lim = lim

Factor out

h from the numerator.

h →0

Cancel the common factor of h .

h →0

=2 x −2

Evaluate the limit.

Find the derivative of f ( x ) = x 2 .

3.6

We use a variety of different notations to express the derivative of a function. In Example 3.12 we showed that if f ( x ) = x 2 −2 x , then f ′( x ) =2 x −2. If we had expressed this function in the form y = x 2 −2 x , we could have expressed the derivative as y ′ =2 x −2 or dy dx =2 x −2. We could have conveyed the same information by writing d dx ⎛ ⎝ x 2 −2 x ⎞ ⎠ =2 x −2. Thus, for the function y = f ( x ), each of the following notations represents the derivative of f ( x ):

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