DERIVATIVES USING THE DEFINITION - Friends University

[Pages:9]DERIVATIVES USING THE DEFINITION

Doing derivatives can be daunting at times, however, they all follow a general rule and can be pretty easy to get the hang of. Let's try an example: Find the derivative of ! ! = !!, and then find what the derivative is as x approaches 0.

The first thing we must do is identify the definition of derivative. The definition is !! ! =

lim!! !

! !!(!).

!!!

In our example, !

!

= !!. (Remember: we can find what !(!) is by

replacing every x in our equation with c). Now that we have our definition, let's find the

derivative. Note: we must first simplify the equation down as far as possible before we take the

limit, or else it may skew our answer.

First we must replace ! ! with our function:

! ! -! !

!! - !!

lim

= lim

!! ! - !

!! ! - !

Now we need to reduce our equation as much as possible. Since we are taking the limit of the

function, we are allowed to do this:

!+! !-!

= lim

!! ! - !

= lim ! + !

!!

Now we take the limit of the function at c: = ! + ! = 2!

Now we are asked to find what derivative is as x approaches 0. For this, we need to now replace every c in our solution with 0.

2! = 2 0 = 0

Therefore, our derivative is 2! and equals 0 as x approaches 0.

Note: there is another accepted definition of derivative which is !! !

=

lim!!

!

!!! !!(!).

!

If

we let ! ! = !!, then:

!! !

= lim

! + ! - !!

!!

!! + 2! + ! - !!

= lim

!!

2! + !

= lim

!!

= lim 2! + = 2!

!!

We would then replace x with 0 to find our derivative as x approaches 0.

2! = 2 0 = 0

Notice, both definitions gave us the same solution.

For a video of this, please reference

DERIVATIVES USING THE POWER RULE

Sometimes using the definition of derivative can be quite cumbersome, but luckily there is a shortcut we can use to find the derivative. Let's look at an example:

Determine the derivative of ! ! = !! + 6! - 2 using the power rule.

The first thing we must do is remember what our power rule is:

!" ! ! = !!, !!" !! ! = ! !!!!

We must remember that the derivative of f(x) plus g(x) is the same as the derivative of f(x) plus

the derivative of g(x). That is ! ! ! + ! ! = !" ! + !" ! . So for our example:

!"

!"

!"

!" !

! =

!! + 6! - 1

!" !"

! !! ! 6! ! 1

=

+

-

!" !" !"

Now we can take the derivative of each individual piece using the power rule:

? ! !! = 2 !!!! = 2!

!"

?

! !! = 1 6 !!!! = 6!! = 6 *Note: !! is 1.

!"

?

! ! = 0 *Note: the derivative of all constants is zero.

!"

Now we can solve our problem:

!! ! = 2! + 6 + 0 ! ! = 2! + 6

For a video on this, please reference

DERIVATIVES USING THE PRODUCT RULE

Sometimes using the definition for finding a derivative can be cumbersome, especially when there are multiple functions of x combined together. However, we have the product rule to help us out. Let's look at an example: Determine the derivative of ! ! = !!!. The first thing we must do is remember the product rule:

! !" = !!! + !" !" Now we must define what u and v are and find their derivatives (we are allowed to use the power rule here): ! = !, !! = 1 ! = !!, !! = !! Now all we have to do is replace our values and simplify: ! !!! = !!! + 1!! !" = !!(! + 1) *Note: This can be expanded to include more than two variables, i.e. ! !"# = !"!! + !!!! + !!" !" For a video on this, please reference

DERIVATIVES USING THE QUOTIENT RULE

Sometimes using the definition for finding a derivative can be cumbersome, especially when there are fractions involved. However, we have the quotient rule to help us out. Let's look at an example: Determine the derivative of ! ! = !!!!. The first thing we must do is remember the quotient rule:

! ! !!! - !" !" ! = !!

Now we must determine which part is u, and which part is v. (Note: when using the product rule, it didn't matter which one was u and which one was v. When using the quotient rule it is imperative that the top be defined as u and the bottom be defined as v):

! = !!, ! = !!,

!! = !! !! = 2!

Now all we have to do is substitute everything:

! !! !!!! - 2!!!

!" !! =

!! !

Now all we have to do is use algebra to simplify:

!!!! - 2!!! !!!(! - 2)

!! !

=

!!

For a video on this, please reference

DERIVATIVES USING THE CHAIN RULE

Sometimes expanding an equation just to take the derivative can be long and drawn out, which leaves us far too many opportunities to make mistakes. Luckily, we have the chain rule to help us out. Let's look at an example: Find the derivative of ! ! = !! - 3! !. We could foil this out, but that would take a very long time. Let's use the chain rule, which is:

!" ! ! = ! ! , !!" !! ! = ! ! ! !! ! For our example:

! ! = !! - 3! !, ! ! = !! - 3! We start by taking the derivative of the outside and leave the inside alone:

! ! ! = 5 !! - 3! ! Now all we have to do is find the derivative of ! ! :

!! ! = 2! - 3 Now we just multiply the two to find our derivative:

!! ! = 5 !! - 3! ! 2! - 3 For a video on this, please reference

IMPLICIT DIFFERENTIATION

Sometimes we need to take the derivative of an equation, but solving for a variable first may be exceedingly difficult. For this we can use implicit differentiation. Let's try an example:

Find the equation of a line that touches !!! + !!! + !" + ! + ! = 3 at the point 1,2 .

Note: we must remember that the derivative of y not 1, but y'. We also must note that every time we take the derivative of a variable that is not x (or more generally, not our independent variable), we must at the derivative of that variable. So, just like every other equation we take the derivative of, we can isolate each term and take the derivative of each term separately.

?

! !!! = 2!" + !!!

!"

Note: We had to use the product rule here.

?

! !!! = !! + !" !!

!"

Note: After we took the derivative of y, we placed a y'.

?

! !" = ! + !!!

!"

? ! ! =1

!"

?

! ! = !!

!"

? ! 3 =0

!"

Note: We have to take the derivative of everything,

including the values on the right-hand side.

Now we can replace the values:

! !!! + !!! + !" + ! + ! = 3 !" 2!" + !!!! + !! + !" !! + ! + !!! + 1 + !! = 0

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