Question

d/dx x^2-x-1=2x-1? or = 2x-x-1
does it have to have a parenthesis?

Answer 1

-1 is a constant to begin with; or rather, -1x^0

Answer 2

to derive it, follow the power rule

Answer 3

d/dx (x^2-x-1)
or
d/dx (x^2-x) -1

Answer 4

what does the "have to have ..." part mean?

Answer 5

the problem says \[d/dx x^2-x-1 \] is it the same as \[d/dx(x^2-x-1)\]??

Answer 6

first one is not clear

Answer 7

\[\frac{d}{dx}(1x^2 -x -1)\]
\[\frac{d}{dx}(1x^2)- \frac{d}{dx}(1x^1)- \frac{d}{dx}(1x^0)\]
power rule says to bring down the exponent and subtract 1
\[\frac{dx}{dx}2*1x^{2-1}- \frac{dx}{dx}(1*1x^{1-1})- \frac{dx}{dx}(0*1x^{0-1})\]
\[\frac{dx}{dx}2x^{1}- \frac{dx}{dx}(1x^{0})- \frac{dx}{dx}(0x^{-1})\]
and dx/dx=1
\[2x-1\]