Differentiation Rules

 Constant Rule and Power Rule

We have seen the following derivatives:

1. If f(x) = c, then f '(x) = 0
   

2. If f(x) = x, then f '(x) = 1
   

3. If f(x) = x², then f '(x) = 2x
   

4. If f(x) = x³, then f '(x) = 3x²
   

5. If f(x) = x⁴, then f '(x) = 4x³
   

This leads us the guess the following theorem.

Theorem d/dx (xn) = nxn-1

Proof:

First note that the binomial theorem says that 

        (x + h)ⁿ  =  xⁿ + nhx^n-1 + _nC₂h²x^n-2 + ... + nh^n-1x + hⁿ 

        =   xⁿ + nhx^n-1 + h²(Stuff) 

For our purposes, stuff is some polynomial in h and k that we do not care about.

We have

               

Sum Difference and Constant Multiple Rules

Theorem:  If f and g are differentiable then

A)   [f + g]' = f ' + g'

B)  [f - g]' = f ' - g'

C)  [cf]' = c(f ')

Proof of C)

                          cf(x + h) - cf(x)
            lim                                          
             h ^->0              h

                             c[f(x + h) - f(x)]
            =  lim                                          
                 h ^->0              h

                                f(x + h) - f(x)
            =  c lim                                      
                   h ^->0              h

            =  c f '(x)

Example  

We know that if f(x) = x² then f '(x) = 2x and that if f(x) = x then f '(x) = 1 so that if

        h(x) = 4x² -3x

then

        h'(x)  =  4(2x) - 3(1)  =  8x - 1

More Applications

Example

Find the derivatives of the following functions:

1. f(x) = 4x³ -2x¹⁰⁰
   

2. f(x) = 3x⁵ + 4x⁸ - x + 2
   

3.  f(x) = (x³ - 2)²
   

4. f(x)  =  10
   

5. f(x)  =  6 / x³ 
   

Solution  

We use our new derivative rules to find

1. 12x² - 200x⁹⁹
   

2. 15x³+32x⁷-1
   

3. First we FOIL to get
   
           [x⁶ - 4x³ + 4]' 
   
   Now use the derivative rule for powers
   
           6x⁵ - 12x² 
   

4. Begin by writing the square root sign as a fractional exponent
   
           10x^1/2 
   
   Now use the power rule to get
   
           f '(x)  =  10(1/2)x ^-1/2  =  5x ^-1/2

Example:

Find the equation to the tangent line to 

        y = 3x³ - x + 4 

at the point (1,6)

Solution:

        y' = 9x² - 1 

at x = 1 this is 8. Using the point-slope equation for the line gives

        y - 6 = 8(x - 1) 

or 

        y = 8x - 2

Example:

Find the points where the tangent line to 

        y = x³ - 3x² - 24x + 3 

is horizontal.

Solution:

We find 

        y' = 3x² - 6x - 24

The tangent line will be horizontal when its slope is zero, that is, the derivative is zero.  Setting the derivative equal to zero gives:

        3x² - 6x - 24 = 0 

or

        x² - 2x - 8 = 0 

or

        (x - 4)(x + 2) = 0

so that 

        x = 4 or x = -2

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