# Question on definition of a derivative

1. Feb 14, 2010

### cue928

1. The problem statement, all variables and given/known data
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)

2. Relevant equations
lim h->0 f(a+h) - f(a) / h

3. The attempt at a solution
Nothing even close to the answer.

2. Feb 14, 2010

### tiny-tim

Hi cue928! Welcome to PF!

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

Show us what you've tried, and where you're stuck, and then we'll know how to help!

3. Feb 14, 2010

### cue928

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. Feb 14, 2010

### vela

Staff Emeritus
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. Feb 14, 2010

### rsa58

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 seperate 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. Feb 14, 2010

### rsa58

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. Feb 14, 2010

### cue928

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. Feb 14, 2010

### vela

Staff Emeritus
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. Feb 14, 2010

### cue928

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. Feb 14, 2010

### vela

Staff Emeritus
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).