Question on definition of a derivative

In summary, Homework Equations Cue928 was not able to find the derivative of f(x) = 1/(x-2) + (x-1)^(1/2) using the definition of the derivative. He tried grouping the x cubed and a cubed together and factoring but he does not know where he is supposed to go. He tried using the X2 tag just above the Reply box but he still does not understand how to use it. He needs more details in order to proceed.
  • #1
cue928
130
0

Homework Statement


I am being asked to find the derivative of f(x) = 1/(x-2) + (x-1)^(1/2) using the definition of the derivative but I am not coming up with anything close (I know what the actual derivative is)


Homework Equations


lim h->0 f(a+h) - f(a) / h


The Attempt at a Solution


Nothing even close to the answer.
 
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  • #2
Hi cue928! Welcome to PF! :smile:

(try using the X2 tag just above the Reply box :wink:)

Show us what you've tried, and where you're stuck, and then we'll know how to help! :smile:
 
  • #3
Ok so the full question is find the slope of the tangent line to the curve y=x^3 - x at (1,0) using lim x->a of f(x)-f(a) / x-a

I had set it up as: (x^3 - x - a^3 + a)/(x-a). I had tried grouping the x cubed and a cubed together and factoring but I don't know where I'm supposed to go. I know the derivative is 3x^2 - 1 and the slope of the tangent is 2 but I apparently don't understand how to use this version of the definition.
 
  • #4
What did you get on top after you factored the x^3-a^3? That is the approach you want to take, but you don't tell us much when all you say is you tried it and got stuck. We need details.
 
  • #5
both of these questions have to do with factorization of the expression that you get from the definition of the derivative. when you apply the formula you need to get rid of that factor in the denominator which is either h or x-a. for the first one separate it into the sum of two derivatives, the first derivative is gotten by collapsing the difference of the two fractions into one which will have a factor of h in both the numerator and denominator. similiarly the next derivative in the sum is gotten by applying the definition but this time multiplying the denominator by the CONJUGATE of the the square roots. evaluating at h=0 gives you the answer. I'm not sure if its called the conjugate but it performs this function of eliminating h in num and denom. the last question is the same but it involves another little trick which as you said is factoring out the x-a. then you evaluate at a and you will get your answer. use the forumla for the difference of two like powers.
 
  • #6
here is a solution to the easiest question: (x^3 -x -a^3 +a)= ((x^3 - a^3) -(x-a))=
((x-a)(x^2 +xa + a^2) - x-a) = (x-a)(x^2 +xa + a^2 -1). now just cancel the factor and plug in the value a. 3a^2 -1 is the familiar answer.
 
  • #7
Thanks for your replies. After I factor, I get (x-a)(x^2+ax-a^2)/(x-a). Cancel the x-a's and I'm left with x^2+ax-a^2. Where do I go from there? I think that's the problem - I'm not sure where I'm supposed to end up.
 
  • #8
Just take the limit now. By the way, you have a sign mistake, and don't forget the other x-a term in the numerator.
 
  • #9
Yeah what I meant to type was:
[(x-a)(x^2+ax-a^2)(a-x) ]/(x-a)
No sign error there unless I'm missing something obvious.
 
  • #10
If you multiply it out, the a^3 term will be positive, but it's negative in the initial expression.

Perhaps just another typo, but you should have a plus sign between the first two factors and the (a-x).
 

1. What is the definition of a derivative?

The definition of a derivative is the rate of change of a function at a specific point. It is represented as the slope of the tangent line at that point.

2. How is the derivative of a function calculated?

The derivative of a function is calculated using the limit definition, which involves taking the limit of the difference quotient as the change in x approaches 0.

3. What is the purpose of finding the derivative of a function?

The derivative of a function helps us understand the behavior of the function and its rate of change at different points. It is an important concept in calculus and is used in various fields such as physics, engineering, and economics.

4. Can the derivative of a function be negative?

Yes, the derivative of a function can be negative. This indicates that the function is decreasing at that point. A positive derivative indicates that the function is increasing at that point, and a derivative of 0 indicates a horizontal tangent line.

5. Is the derivative of a function always defined?

No, the derivative of a function may not always be defined. It is not defined at points where the function is discontinuous or has sharp corners. It is also not defined at points where the function is undefined or has a vertical tangent line.

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