f(x)=2x^2 - 1
I got to:
f'(x)=lim z->x (2z^2 - 2x^2 - 2)/(z-x)
Problem is, I can't seem to factor (z-x) out of the numerator. I was just wondering if I had missed something or done my derivative wrong. Any help would be much appreciated.
Name:
Anonymous2009-03-18 19:10
that -2 should be 0.
f(z)-f(x)= (2z^2 - 1) - (2x^2 - 1) = 2z^2 - 1 - 2x^2 + 1
When your variable is raised to a power of n, the derivative is pretty simple. a is constant.
d_a*x^n = a*n*x^(n-1) dx
fuck all of that other stuff. learn your derivative rules.
Name:
Anonymous2009-03-18 21:30
>>3
Don't worry about it, I mess up signs all the time. I've just learned to go back and check them first if something isn't turning out correctly.
>>6
The point of using the basic definition of a derivative is so that we know why we can use those derivative rules. If you want to learn higher level math, it is important to understand why something is applicable first rather than blindly using rules like that.
Name:
Anonymous2009-03-19 22:21
use the other limit definition of derivative:
lim[(f(x+h) - f(x))/h, h->0]
way easier in most situations
Name:
Anonymous2009-03-20 16:53
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