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Determine the slope of the line tangent to ƒ(x)=2x-3 at point (x,y).

Solution by the four-step process,

1)           ƒ(x+x) = 2(x+x)-3

2)   ƒ(x+x)-ƒ(x) = (2x+2x-3)-(2x-3)

                              = 2x

3)

           

4)

           

Note in Example 1 that the slope of ƒ is constant. This is not surprising since we know that the graph of ƒ is a line. Of course, not all graphs have constant slope, as the next example illustrates.


Determine the formula for the slope of the graph of y = x2 + 3 (see figure 5).
What is its slope at (0,3)? At (-2,7)?



                                                      

Solution by the four-step process,

1)           ƒ(x+x) = (x+x)2+3

2)   ƒ(x+x)-ƒ(x) = x2+2x(x)+(x)2+3-x2-3

                              = 2x(x)+(x)2

3)

           

4)

           



Find the derivatives of ƒ(x)=x2+2x.

Solution

1)           ƒ(x+x) = (x+x)3+2(x+x)

2)   ƒ(x+x)-ƒ(x) = x3+3x2(x)+3x(x)2+(x)3+2x+2(x)-(x3+2x)

                              = 3x2(x)+3x(x)2+(x)3+2(x)

3)

           

4)