© 2009  Rasmus ehf    & Jóhann Ísak

Derivatives

Lesson 4

Derivatives of exponential and trig functions

We're going to use the CASIO-calculator to find some values of functions of the type f(x) = ax  where a is a constant and x is the variable. We will do this by fixing the value of x and looking at values of f(x) and  f´(x)  as a changes from 2 to 3. We can choose any value for x, for example x = 2. Choose the RUN menu and then the button labled  OPTN next to the  SHIFT-button. Next choose CALC with F4 and then d/dx withF2. Finally put in a komma and then the x value 2.

 

The results are shown in the table below

The value of  a² goes from  4 when a = 2 up to 9 when a = 3. The derivative goes from  2.8 up to  9.9 (both numbers are approximate). Both columns of numbers change continuously so somewhere, not far from 2.7, the function and it's derivative take the same value.
We will see this happening in exactly the same place whatever value we choose for a.
If we choose  a = 2.718282 we get the same value in both columns up to the sixth decimal place (7.3890570 and 7.3890575). You should recognise this number from the lesson on natural logarithms. It's the number e.

The number e is an irrational number and therefore we can only give an approximate value for e, e 2.718282 . The function f(x) = e^x remains unchanged when we differentiate it, that is f(x) = f´(x) = e^x 

In lesson 5 you will see a rule called  the Chain Rule. One of the results of this rule is that if x is multiplied by a constant number then the differentiated function is also multiplied by that constant . For example if we differentiate f(x) = e^2x  we get  f´(x) = 2e^2x .

In general, if k is a constant and f(x) = e^kx  then  f´(x) = ke^kx .

Using this fact we can find the derivative of the function f(x) = a^x. Using the log rules x = e^{ln x} and ln a^x = x·ln a  we can rewrite f(x) as 

   f(x) = a^x =   e^{x·ln a}

a is a constant, so ln a is also a constant like the k in the above rule. So we can differentiate the function by writing it in the form e^{x·ln a}.

   f´(x) = (e^{x·ln a})´ = (ln a)·e^{x·ln a} = (ln a)·a^x

Example 1

Find the derivative of f(x) = e^x·a^x.

Using the multiplication rule (uv)´= u´v + uv´ with u = e^x and v = a^x gives us u´ = e^x and v´ = (ln a)·a^x. Putting these values into the formula we get

   f´(x) = (uv)´= u´v + uv´

           = e^x·a^x + e^x·(ln a)·a^x        take  e^xa^x outside a bracket .

           = e^xa^x(1 + ln a)

Now we consider a rule  often referred to as the sandwich rule.  This is a method we can use to solve difficult limit problems.  If we are unable to find the limit of a function  f(x) in a particular point P we can try to find a function u(x) that has the same value as f(x) in the point P but is greater than f(x) in the neighbourhood of P. In the same way we find a function  v(x) that takes the same value as f(x) in the point P but is less than f(x) in the neighbourhood of P. In this way we sandwich f(x) in between
u(x) and v(x) . If these two functions u(x) and v(x) have the same limit in point P then f(x) must also have this limit. This is demonstrated in the diagram below.

Now we will use the sandwich rule to find  . The problem is that if we put  0 in for the x then we get 0/0 which we have a problem evaluating.

The above diagram shows the unit circle. We have drawn the angle x. If x is measured in radians then the length of the arc  between the point (1, 0) on the x axis and the point P is also x . The vertical red line has length  sin x and is obviously less than the arc x which is again less than tan x . We can write the following inequality:

   sin x x tan x

Remember  tan x is defined as  so we get

Dividing all through by  sin x and cancelling

Inverting all the fractions and reversing the inequality symbols gives us

We have succeeded in sandwiching the fraction    between 1 and cos x
which both have the limit 1 as x tends to 0 .

  This gives us the rule     = cos 0 = 1

We use this rule in the following examples .

Consider the limit  .

In this case we don't need to use the sandwich rule, instead we use  (a + b)(a – b) = a² – b² first multiplying the fraction all through by  (cos x + 1)

Remember Pythagoras's rule for cos and sin,  cos²x + sin²x = 1 which can be rewritten as  –sin²x = cos²x –1.

We continue the example as follows:

The first fraction has the limit 1 from the above rule and the second fraction is 0 because sin 0 = 0. This gives the result

Example 2

We can now set about finding the derivative of  sin x.

(sin x)' =

Using one of the addition rule for trig functions  sin(u + v) = sin u cos v + cos u sin v and putting it into the above expression we get:

In the last line of the proof we used the results of the two limit rules that we proved earlier in this lesson. We get the very satisfactory result that the derivative of sin x is cos x.

A similar method gives the result that the derivative of cos x is – sin x.

Example 3

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

Remember that  and use the rule for the derivative of quotients  . 

In this case u = sin x and u´ = cos x,  v = cos x and v´= – sin x. Putting these values into the rule we get

Here we use the rule cos2x  + sin2x = 1.

Practise these methods and then take test 4 on derivatives.

ps. Remember your check list.