Question

find the derivative of the function by using the definition of the derivative
f(x)=2/x

Answer 1

Start by using:\[\lim_{h \rightarrow 0} [2/(x + h) - 2/x] / h\]

Answer 2

Then, for the expression between the brackets, use (x + h)x as a common denominator. The "h" will cancel out.

Answer 3

so it would be 2(x+h)-2(x)/h

Answer 4

and it would be 4x^2

Answer 5

Between the brackets, and just looking at that portion for now, 2/(x+h) - 2/x = [2x - 2(x+h)] / [x(x+h)]

Answer 6

So, first simplify (after seeing how I got that equation) and plug that into the limit expression inside the brackets.

Answer 7

so I'd put 0 for it

Answer 8

Are you saying that 2x - 2(x+h) is 0?

Answer 9

yeah…?

Answer 10

2x - 2(x+h) = 2x - 2x - 2h. And that does not equal 0. What does it equal?

Answer 11

-2?

Answer 12

2x - 2(x+h) = 2x - 2x - 2h = (2x - 2x) - 2h = 0 - 2h. What does that equal?

Answer 13

0-2=-2

Answer 14

You are disregarding the "h". Look again. I did not write 0 - 2. I am asking you what 0 - 2h equals.

Answer 15

-2h???? ahhh im confused

Answer 16

It's OK to be confused, as long as you can "see the light at the end of the tunnel". Many sources of math confusion are just little hurdles that, once over, you never stumble on again. Just think of "h" as another variable, similar to "x" or "y", that's all. And yes, you are making progress. In the solution to this problem, this intermediate answer is -2h so far. But we're not done yet.

Answer 17

So, the "between the brackets" expression, 2/(x+h) - 2/x = [2x - 2(x+h)] / [x(x+h)] now equals -2h / [x(x+h)]. So now, you can put that into the limit expression in my first post.

Answer 18

So, you should have\[\lim_{h \rightarrow 0} ([-2h] / [x(x + h)]) / h\]

Answer 19

I dont understand where the their 2 came from

Answer 20

third*

Answer 21

This simplifies to lim h→0 (−2h)/[x(x+h)h]. So, divide out the factor of "h" now (NOT the "x+h").

Answer 22

nvm I goyt it

Answer 23

Okay, good so far. After you divide out the "h" factor, you've got lim h→0 (−2)/[x(x+h)]. So, now the h can go to zero without giving "0" in the denominator.

Answer 24

The denominator of x(x+h) will become x(x+0) or x(x) which is x^2

Answer 25

And since the numerator is just -2, you put numerator and denominator together for your answer and you're done!

Answer 26

h is replace by 0 so it disappears?

Answer 27

"h" disappears only at the end when you take the limit. "h" stays as "h" up until the last step (taking the limit) all while you are doing the arithmetic and the simplifying. For most of the problem, "h" is not 0 and is just a variable.

Answer 28

A good way to conceptualize "h" is "change in x" or "delta x" or :\[\Delta x\]

Answer 29

right. ok I think I got this now. Thankyou so much for your help, and for taking the time to explain this tome . YOu were a huge help. thanks again..

Answer 30

Starting to make a little sense? I could give you a little more mathematical theory if that would help. With some students, that helps. Others just want to know the "how to get the answer" part.

Answer 31

how about finding the slope of A tangent line..?

Answer 32

Once you know how to get the derivative that we just did, the slope of a tangent line is a real snap. It's just the derivative! And if you need a concrete number at a certain value of x, you just plug that value into the x and you have the slope "at that point". If you want the slope in general, then just leave the "x" as "x".

Answer 33

The derivative for a function is itself just another function. And you could take the derivative of that derivative function. It would end up being the second derivative of the original function.