How do you find the derivative of #f(x)=2x^3-x^2+3x#?

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Answer 1 · 2017-03-02T18:52:13

#6x^2-2x+3#

Derivative of any simple power is

#d/dx ax^n = n*ax^(n-1)#

and for polynomials,

#d/dx (a+b+c)=d/dxa + d/dxb + d/dxc#.

For the above question,

#f(x)=2x^3-x^2+3x#

#d/dx(2x^3) = 6x^2#

#d/dx(-x^2) = -2x^1=-2x#

#d/dx(3x)=3x^0=3#

so

#f'(x)=6x^2-2x+3#