The Fug is above  Derivatives
literally

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Differentiation

Without too much attention to detail and theory, we have five basic ways to differentiate
and two rules to follow.

• First Way         Simple Differentiation
• Second Way     Chain Rule Differentiation
• Third Way      Logarithm Differentiation
• Fourth Way     Exponential Differentiation
• Fifth Way       Implicit Differentiation
• First Rule         Product Rule
• Second Rule      Quotient Rule

First Way     Simple Differentiation

This is called the power rule in most books.

This is how you go:

• multiply by the power and take one away from the power
• Power(x) ^{( power - 1 )}

So let’s go through the examples:

Example One

f(x) =  3 x²¹
f ’(x) = 21 ´ 3  x ^{21 - 1}       =     63  x ²⁰

The answer is    f ’(x) = 63  x ²⁰

Notice if you have a number there already just times the number by the power.
Watch out for  x ^1/2 and x  ^{- 1}, it is those negative numbers again!!

f( x)  =   x ^1/2
f ’(x)  = 1/2  x ^{-  1/2} =     1    
                                   2Ö  x

The answer is        f ’(x) =    1    
                                           2Öx

f( x)  =   x ^{-  1}
f ’(x)  = -  1  x ^{-  2}   =   -  1
                                       x²

The answer is        f ’(x) = -  1/ x²

Back to top         Back to Product Rule           Back to Quotient rule

Second Way     Chain Rule Differentiation

This is where we have a slightly more complicated looking function than the simple
differentiation function. 
This method I call “The twist” as you use the simple method with a twist at the end.

The function looks more complicated such as 3(x³ + 21 -  10x)⁷ or (x + 4x³)^1/2

This is how you go: 

• Differentiate as if it is simple, multiply by the power and take one away from power.
• Because it is not simple, here is the twist multiply by the derivative of the bracket.
• Simplify if possible

So let’s go through the examples

Example Two

f (x) = 3(x³ + 21 -  10x) ⁷
f ’(x) =  (7 ´  3)                   (x³ + 21 -  10x) ⁶                  (3x²-  10)

(multiply by power)   (Original bracket) ^{Power-  1}(derivative of the bracket)
 

The answer is      f’(x) =  21 (3x²-  10) (x³ + 21 -  10x) ⁶

Example Three

f (x) =    (x + 4x³) ^1/2
f ’(x) =    1/2                         (x + 4x³)  ^{-  1/2}                     (1 + 12x²)

  (multiply by power)   (Original bracket) ^{Power-  1}(derivative of the bracket)
                                                                                   (Using Simple differentiation)

 

The answer is    f ’(x) =     1/2 (1 + 12x ²) (x + 4x ³) ^{-  1/2}

 Back to top         Back to Product Rule

Third Way     Logarithm Differentiation

I find it handy to put brackets around my x-values to help when differentiating.

This is how you go:

• Put bracket around the expression being logged.
• Use logarithm rules to simplify if possible.
• The answer is a fraction.
• The denominator is the original bracket.
• The numerator is the derivative of the bracket.

So let’s go through the example:

Example Four

f (x) =  ln x ³ +5x

f (x) =  ln (x ³ +5x)

f ’(x) =  3x ² + 5     Derivative of the bracket (using simple differentiation in this case)
             x ³ +5x           Original Bracket

The answer is      f ’(x) =  3x ² + 5          
                                         x ³ +5x

Back to top         Back to Product Rule         Back to Quotient rule

 

Fourth Way     Exponential  Differentiation

This differentiation always has the original function in the answer, once again I find it
handy to put brackets around the power.

This is how you go:

• Put brackets around the power.
• Write down the original exponential.
• Multiply by the derivative of the bracket.

f(x) = e^(bracket)

f’(x) ⁼ e^(bracket) . ( derivative of the bracket).

So let’s go through the example:

Example Five

f(x) = e (-  3x)

f ’(x) =  e ^{(-  3x)}    .     (-  3 )

f ’(x) = e ^(bracket)  .( derivative of the bracket).

The answer is      f ’(x) = -  3 e ^{-  3x}

Back to top         Back to Product Rule

Fifth Way     Implicit Differentiation

This is not difficult and in one way is similar to the chain differentiation.  We write down
dy/dx or f ’(x) without saying or thinking that differentiating y or f(x) gives us dy/dx or  f ’(x)
as we are so worried about differentiating the x-values on the other side.  This time we have
not got “ y = some expression” instead we have the x and y values mixed together.

This is how you go:

• Check to see if you can get the expression into y = ......
• Differentiate the x-values as normal.
• Differentiate the y-values as if they are x-values.
• But the twist is any y-value must be multiplied by dy/dx.
• Move all terms with dy/dx to one side, all other terms on the other side.
• Take dy/dx out as a factor and divide by the bracket.

So let’s go through the example

Example Six

y ² + 4y + 7 = e ^6x
 

Now we cannot simplify the expression to y = .....   so follow the instructions

         2 y ^{(2 - 1)}         dy/dx           +         4                   dy/dx         +   0       =      6 e^6x

(Treat just        (multiply by       (Treat just        (multiply by                 = (Exponential
like x-value)        dy/dx)           like x-value)        dy/dx)                         Differentiation)

2y dy/dx + 4 dy/dx  = 6 e ^6x


dy/dx  ( 2y + 4 )  =   6 e ^6x     (Take dy/dx out as a factor)
 

dy/dx      =       6 e ^6x               ( Divide by the bracket)
                       ( 2y + 4 )
 

The answer is      f ’(x) =      6 e ^6x            
                                                   ( 2y + 4 )

Back to top

 

First Rule       Product Rule

This is a case of following an easy procedure and substituting values into the expression.
The Product Rule can involve all the different types of differentiation

This is how you go:

• Put brackets around the two terms (it could be more but not usually).
• Differentiate the terms separately.
• Multiply the first term by the derivative of the second term
• Add the answer above to the second term multiplied by the derivative of the first term.
• This is the answer.

If u = first term                       then  u’ = derivative of first term and
   v = second term                          v’ = derivative of second term

The derivative of u v      =       u   v’    + v  u’

So let’s go through the example,

This example has a few of the differentiations not all products are as complicated
but it is good to see them before a test and not in the test for the first time.

Example Seven

f(x) =( e ^3x)(x ² + ln x) ³

Let u =  e ^3x                       v = (x² + ln x) ³

     u’ = 3e ^3x                     v’ = 3(x ²+ ln x ) ² ( 2x    +     1/x)

   Exponential                             Chain       Simple     Logarithm

Now you can substitute the values into the expression    u  v’   +  v   u’ .

If you set the answer out as above, you may just multiply diagonally
and add the answers. This is exactly the same as the expression u  v’   +  v   u’ .

d(u v) / dx is the mathematical way of writing the derivative of a product

     d(u v) / dx          =    u                        v’                      +           v                u’

d(e ^3x(x ² + ln x) ³) / dx   =    e ^3x     3(x ²+ ln x ) ²( 2x +1/x)   +   (x ² + ln x) ³      3e ^3x
 

Take out any factors, I will just write f ’(x) for d(e ^3x(x ² + ln x) ³) / dx   so

f ’(x) = 3e ^3x (x ²+ ln x ) ²  { 2x + 1/x + (x ² + ln x) }
 

We will leave that as the answer but you can always do more simplification if you want.

The answer is      f ’(x) = 3e ^3x (x ²+ ln x ) ²  { 2x + 1/x + (x ² + ln x) }

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Second Rule       Quotient Rule

Again, this is a case of following an easy procedure and substituting values into the expression. 
The Quotient Rule can involve all the different types of differentiation.

This is how you go:

• Differentiate the terms separately
• Multiply the bottom term by the derivative of the top term
• Subtract from the answer above the top term multiplied by the derivative of the bottom term.
• Divide the whole of this subtraction by the square of the bottom term

If v = bottom term                          v’ = derivative of bottom term
   u = top term                               u’ = derivative of top term

The derivative of u /  v     =   v  u’   -   u  v’    
                                         v²

So let’s go through the example:

Example Eight

f(x) =   ln (3x ² + 4)
                     x ³

Let    v = x ³                        u = ln (3x ² + 4)

         v’ = 3x ²                      u’ =     6x     
                                               3x ² + 4

Simple                           Logarithm

Now you can substitute the values into the expression      v  u’   -   u  v’
                                                                                                       v ²

If you set the answer out as above, starting from top left, you may just multiply diagonally
and subtract the answers and divide by the square of the bottom. 
This is exactly the same as the expression       v  u’   -   u  v’
                                                                                   v ²

d(u /  v) / dx is the mathematical way of writing the derivative of a quotient

     d(u /  v) / dx                       =       v               u’               -         u                     v’

     d(ln(3x²+ 4) /x³) / dx       =       x ³      6x / (3x ² + 4)     -     ln (3x ² + 4)        3x ²
                                                                                                                                        

               ( x ³) ²

                   v ²

Simplify as much as possible:

f ’(x)  = 6x⁴ / (3x² + 4)     -     3x² ln (3x²+ 4)      
                                                                            

x⁶
 

To make it easy just take the top by itself and making it one term:

f ’(x)  =       6x⁴   -     3x²⁽ ln (3x² + 4)       
             (3x² + 4)

 

=      6x⁴         -     3x²(3x² + 4) ln (3x² + 4)       
                                                                      

                 (3x² + 4)

 

=      3x²  {2x²   - (3x² + 4) ln (3x² + 4)}      
                                                                              

                 (3x² + 4)

Putting the bottom back in,  x⁶
 

=      3x²  {2x²   - (3x² + 4) ln (3x² + 4)}      
                                                                              

                 x⁶ (3x² + 4)

Cancelling x², top and bottom
 

=      3{2x²   - (3x² + 4) ln (3x² + 4)}      
                                                                              

             x⁴(3x² + 4)

 

The answer is      f ’(x)  =   3{2x²   - (3x² + 4) ln (3x² + 4)}      

x⁴(3x² + 4)    

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