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find f prime of x if f(x) = x^2-x I understand that f prime (x)= 2x i just dont understand how to work the -x
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\[f(x) = x^2-x\]\[f'(x) = \text{derivative of \(x^2\) - derivative of \(x\)}\]
Exactly. The derivative of x2 is 2x how do i find out the derivative of x
\[x = x^1\]
\[x^0 = 1, \,x\ne 0\]
Going back to the limit definition of the derivative, \[\frac{d}{dx}[x] = \lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} = \lim_{h\rightarrow 0}\frac{x+h-x}{h} = \lim_{h\rightarrow 0}\frac{h}{h} = 1\]
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so the answer is 2x-1 or just 2x?
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