Differentiation

Basic Differentiation Techniques

Differentiation is a calculus method used to determine the slopes of tangents to a function at any point on the function. These results have significance when considering real world problems. A common example is the relationship between displacement, velocity and acceleration functions of any moving body.

A summary of useful shortcuts for differentiation:

  1. f(x) = a, then f'(x) = 0
  2. f(x) = x, then f'(x) = 1
  3. f(x) = x^a, then f'(x) = ax^{a-1}
  4. f(x) = ln(x), then f'(x) = \frac{1}{x}
  5. f(x) = ln(f(x)), then f'(x) = \frac{f'(x)}{f(x)}
  6. f(x) = e^x, then f'(x) = e^x
  7. f(x) = e^{ax}, then f'(x) = ae^{ax}

These shortcuts have been derived from the first principles approach to differentiation.

Examples


Example #1: Differentiate x^3
Using technique #3, we get:

f'(x) = 3x^{3-1} = 3x^2

Example #2: Differentiate \sqrt{2x}
The function can be re-written like so:

\sqrt{2x} = (2x^{\frac{1}{2}})

Using technique #3, we get:

\frac{1}{2}2{x}^{\frac{-1}{2}}
= x^{\frac{-1}{2}}
= \frac{1}{\sqrt{x}}

Example #3: Differentiate ln(x^{2})
Using technique #5, we get:

\frac{f'(x)}{f(x)} = \frac{2x}{x^2}
Where by using technique #3, we get:

f(x) = x^{2}
f'(x) = 2x

Example #4: Differentiate e^{3x}
Using technique #7, we get:

f'(x) = 3e^{3x}

Activities

 

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Product and Quotient Rules

The product rule and quotient rule are used when differentiating products or divisions of two or more functions.

The Product Rule

If          y = f(x)g(x)
then   y' = f'(x)g(x) + f(x)g'(x)

Notice that our two functions f(x) and g(x) are multiplied together. Our final differentiation of y involves the separate derivatives of these two functions.

Examples

Example #1: Differentiate x^2\sqrt{x}

Determining the two functions:

f(x) = x^2
g(x) = \sqrt{x}

We now need to determine:

f'(x) = 2x

and,

g'(x)=\frac{1}{2\sqrt{x}}

Now by using the rule:

y' = 2x(\sqrt{x})+x^2(\frac{1}{2\sqrt{x}})

No further simplification is necessary, however it may be helpful for more involved questions.

 

The Quotient Rule

If          y = \frac{f(x)}{g(x)}
then   y' = \frac{f'(x)g(x)-f(x)g'(x)}{g(x)^{2}}

In this case, our two functions f(x) and g(x) are being divided. Similarly to the product rule, our final differentiation of y involves the derivatives of f(x) and g(x).

Examples

Example #1: Differentiate \frac{x^2}{2x+3}

Determining the two functions and their derivatives:

f(x) = x^2

g(x) = 2x+3

f'(x) = 2x

g'(x) = 2

Now to apply the quotient rule:

y' = \frac{2x(2x+3)-x^2(2)}{(2x+3)^2}

 

Activities


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The Chain Rule

The chain rule is used for functions which contain other functions. We refer to such functions as being composite.

Composite functions are noted as:

y = f(g(x))

Where f(x) and g(x) are two functions of x. Our variable is first evaluated within g(x) and the resulting value is then evaluated within f(x). Note that the raw value for x changes after each evaluation depending on the functions.

To differentiate such functions we do the following:

y' = f'(g(x))g'(x)

Although confusing, when broken down the chain rule is relatively straightforward. As above, the derivative of a composite function involves the separate derivatives of our two functions f(x) and g(x).

The secret to deriving functions in this way is being able to recognize which functions are which in an equation. In most chain rule cases, there will be an ‘outer’ function and an ’embedded’ function within it.

Examples

Example #1: Differentiate (x^3+1)^2
We first determine which function is which:

f(x) = x^2 as it is the ‘outer’ function.
g(x) = x^3+1 as it is the ’embedded’ function within the outer function.

Note that in our original function, we have f(g(x)) with the function values as above.

Now to differentiate:

f'(x) = 2x
g'(x) = 3x^2

And to apply the chain rule:

f'(g(x))g'(x) = 2(x^3+1)(3x^2)

Example #2: Differentiate \frac{1}{3x^2+2x+5}
Note that this function can be re-written as:

(3x^2+2x+5)^{-1}

It is now in a form where we can apply the chain rule.

Our ‘outer’ function is:

f(x) = x^{-1}

And our ’embedded’ function is:

g(x) = 3x^2+2x+5

Note again the relationship between our two sub-functions and the original function.

To differentiate:

f'(x) = -x^{-2}
g'(x) = 6x+2

And apply the chain rule:

f'(g(x))g'(x) = -(3x^2+2x+5)^{-2}(6x+2)

Note that this question could have also been completed by using the quotient rule with f(x) = 1.

Activities

 

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Mixed Differentiation

More complex functions will require a combination of the previous differentiation techniques covered to determine its derivative. Although there is no set formula for approaching these functions, there is a process to answering them that simplifies the question.

The steps to take when faced with a differentiation requiring multiple methods is as follows:

  1. Differentiate term-by-term
    If the function you are facing has multiple terms, focus on each term separately when differentiating.
  2. Determine what type of term it is
    Is the term a quotient? A product? Are there multiple products within the term? Determining what type of term it is will determine what type of differentiation technique is needed.
  3. Determine the sub-functions of the term
    Are there any functions you can separate from each other within the term? For example is there one function being divided by a composite function in a quotient term? Determining what type of sub-functions are present will help when differentiating.
  4. Apply the relevant differentiation techniques
    After determining the type of term and sub-functions of that term, you can begin differentiating.
    First select the appropriate differentiation method for the type of term. Write down the method with the sub-functions substituted in their respective positions and differentiate, remembering to differentiate sub-functions that compose the respective parts of the method.

This is likely to confuse you at first, so we will go through these steps in examples below.

Examples

 

Activities

 

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Questions