Differential calculus, derivatives

[DERRAN]

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

Find f'(x)= 16x - x-2 using first principles.

2. Relevant equations
x
http://img153.imageshack.us/img153/8403/597137697c1f605c7a43d34qz4.png (http://imageshack.us)


3. The attempt at a solution
I used dy/dx and got 2x-3 + 16 but I get something different when I use the formular I attempted several times and I cant get the same answer as the dy/dx.
Please need help!!!

[HallsofIvy]

Do you mean "find f'(x) if f(x)= 16x- x-1"? That's quite different from what you say!

If f(x)= 16x- x-2, then f(x+h)= 16(x+h)- (x+ h)-2

f(x+h)- f(x)= 16(x+ h)- 16x - (x+h)-2+ x-2

The first part of that is just 16x+ 16h- 16x= 16h.

The second part is
-\frac{1}{(x+h)^2}+ \frac{1}{x^2}
= \frac{-x^2}{x^2(x+h)^2}+ \frac{(x+h)^2}{x^2(x+h)^2}
= \frac{-x^2+ x^2+ 2hx+h^2}{x^2(x+h)^2}= \frac{2hx+ h^2}{x^2(x+h)^2}
so
f(x+h)- f(x)= 16h+ \frac{2hx+h^2}{x^2(x+h)^2}
Now, what is (f(x+h)- f(x))/h and what is the limit of that as h goes to 0.

[DERRAN]

From where you left off...
http://img527.imageshack.us/img527/4471/newpicture.th.png (http://img527.imageshack.us/my.php?image=newpicture.png)
As you can see you still don't get 16 + 2x-3 but instead you get 16+2x-2.

Why???:surprised

[lanedance]

you've lost a square from the denominator for no reason during your calc, they should work

[DERRAN]

Thanks sorry about that it works perfectly!