Question

Can someone show me how to find the derivative at a point using the definition? Example: f(x)=x^2-5x+1 at x=2

Answer 1

the definition of derivative is:
\[\lim_{x \rightarrow a}\frac{ f(x)-f(a) }{ x-a }\]
all you have to do is plus that in

Answer 2

Owl is right, you can find many youtube videos that explain this kind of a problem.
For example: http://www.youtube.com/watch?v=vzDYOHETFlo

Answer 3

I am sorry I did not clarify my question. I understand it is a simple matter of plugging in but what do I plug in for (a) ? as x=2 but I have no value for (a)

Answer 4

at x=2 means a=2

Answer 5

Wait, do we need to teach you the concept of limit first?

Answer 6

The concept of limit is usually demonstrated by \(\large\lim\limits_{x\rightarrow1}\frac{x^2-1}{x-1}\)

Answer 7

Thank you! I know how to find the derivative of a function using the definition just needed some clarification on (a) thank you!

Answer 8

(a) is the point where we wnt to find the derivative in.

Answer 9

thus (a)=2 correct?

Answer 10

exactly.

Answer 11

can you give me the answer please for verification ? I want to see if I am doing it correctly by plugging it in! Thank you!

Answer 12

What is your answer?

Answer 13

I just plugged in the numbers so I have (x\[(x^2-5x+1)-(-5)/ 2-2 \] I do not think that is correct that is not my answer but where I am so far

Answer 14

(Proofs will not be shown because of their complicity)
We have the first rule in differentiation: \(\large\dfrac d{dx} x^n=nx^{n-1}\)

Answer 15

Therefore \(\dfrac d{dx}x^2=2x^1=2x\)

Answer 16

that is just the power rule correct?

Answer 17

Yes that is the power rule

Answer 18

Apply the power rule to \(-5x\) and you'll obtain \(-5\).

Answer 19

The derivative of any constant equals zero.

Answer 20

Therefore \(\dfrac d{dx}x^2-5x+1=\dfrac d{dx}x^2-\dfrac d{dx}5x+\dfrac d{dx}1=2x-5\)

Answer 21

Now plug x=2 inside

Answer 22

so for f(x) I plug in 2 before I take the derivative than plug in 2 for f(a) after doing the power rule? and subtract ?

Answer 23

What @Andras is teaching you is to use the FIRST PRINCIPLE to do differentiation

Answer 24

Which is more formal but more complicated

Answer 25

However, I suppose that your school does not require the use of the FIRST PRINCIPLE.

Answer 26

If you do this in an exam, I would surely take a few points away!
Using the definition.... the question states that you cannot use the power rule.
kc-kennylau is right but this method cannot be used for this question

Answer 27

He didn't say we can't use power rule

Answer 28

yes I have to use the definition to find the answer otherwise it would not be as hard!

Answer 29

I apologize but I have to use the definition to find the derivative no other rules can be used
to my knowledge

Answer 30

I have to use f(x)-f(a)/x-a to solve it

Answer 31

sorry

Answer 32

it is alright I was not clear enough

Answer 33

We'll begin by expressing it on the definition form:
\[\lim_{x \rightarrow 2}\frac{ (x ^{2}-5x+1)-5 }{ x-2 }\]
getting rid of the parenthesis and operating with the -5 :
\[\lim_{x \rightarrow 2}\frac{ x^{2}-5x-4 }{ x-2 }\]
factor the numerator and see.

Answer 34

Owl was super quick :)

Answer 35

I'm lazy, I always find a quick way, jk.

Answer 36

I am left with x-9/-2 after factoring an x out

Answer 37

which is not correct

Answer 38

oh, we traduced it into a finite polinomial indetermination, we have to get rid of the denominator, I got an answer of -1.

Answer 39

Thank you @Owlcoffee for helping me out here btw

Answer 40

It is a lot easier to derive the general formula for the derivative, using the definition. And later plug in a=2.
\[\lim_{x \rightarrow a} \frac{ {x^2-5x+1-a^2+5a-1} }{ x-a }\]

Answer 41

The nominator can be factored
x^2-a^2=(x-a)(x+a)
-5x+5a=-5(x-a)
Thus
\[\frac{ {(x-a)(x+a-5)} }{ x-a }\]

Answer 42

Now cancel x-a to get
\[\lim_{x \rightarrow a}=x+a-5\]

Answer 43

And that is at x=a
a+a-5=2a-5

In this question a=2
2*2-5=-1