Find the derivative by using product rule and distribution
pls help quickly and show work

Below

Step-by-step explanation:

First method:

● f(x)= (x^3-2x+1)×(x-3)

● f'(x)= (x^3-2x+1)' ×(x-3) + (x^3-2x+1)×(x-3)'

●f'(x)= (3x^2-2)×(x-3) + (x^3-2x+1) × 1

●f'(x) = 3x^3-9x^2-2x+6 + x^3-2x+1

● f'(x)= 4x^3-9x^2-4x+7

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Second method:

●f(x) = (x^3-2x+1)×(x-3)

●f(x) = x^4-3x^3 -2x^2+6x+x-3

●f(x) = x^4-3x^3-2x^2+7x-3

●f'(x) = 4x^3-9x^2-4x+7

We got the same result using both methods.

what is the starting amount? the question is not clear sorry.

step-by-step explanation:

intensity of the vector is v= √37 ≈ 6.08 units and make angle ∡α ≈ 9.46° with east direction

step-by-step explanation:

required vector is consists of the two componentsvx= 2+4=6 units   and vy= 1 unit and vx ⊥ vywe will use pythagorian theorem to find intensity of the vector vv∧2 = vx∧2 + vy∧2 => v = √vx∧2 + vy∧2 = √6∧2 + 1∧2 = √36+1 = √37 ≈ 6.08 unitsthe angle ∡α between vector and east direction we wil find with tanαtanα = 1/6 => α = arc tanα = arc 1/6 => α ≈ 9.46°

hope this