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

Answer 1

Ezra Adams

#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#

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

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Answer 2

Christopher Agresta

To find the derivative of ( f(x) = 2x^3 - x^2 + 3x ), you can use the power rule of differentiation, which states that the derivative of ( x^n ) with respect to ( x ) is ( nx^{n-1} ). Applying this rule to each term of the function, the derivative of ( f(x) ) is:

[ f'(x) = \frac{d}{dx}(2x^3) - \frac{d}{dx}(x^2) + \frac{d}{dx}(3x) ]

[ f'(x) = 3 \cdot 2x^{3-1} - 2x^{2-1} + 3 \cdot 1x^{1-1} ]

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

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.