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To take the derivative, apply the rule for sum and difference of functions which is `(u+-v)'=u'+-v'` .
So, take the derivative of each term.
`f'(x)=(2x^3)' - (4x^2)' + (3x)'`
For each term, apply the power rule of derivatives which is `(cx^n)'=c*nx^(n-1)` .
`f'(x)= 2*3x^(3-1) - 4*2x^(2-1) + 3x^(1-1)`
`f'(x)=6x^2 - 8x+3`
Hence, the derivative of `f(x)==2x^3-4x^2+3x` is `f'(x)=6x^2 - 8x+3` .
(Note: For #2 and #3, post them as separate questions.)
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