Question

dy/dx of x^3/3-2x^2+3x​

Answer 1

ANSWER....

$y = \frac{ {x}^{3} }{3} - 2 {x}^{2} + 3x \\ \\ \frac{dy}{dx} = \frac{1}{3} \frac{d( {x}^{3} )}{dx} - 2 \frac{d( {x}^{2} )}{dx} + 3 \frac{dx}{dx} \\ \\ \frac{dy}{dx} = \frac{1}{3} ( 3{x}^{2} ) - 2(2x) + 3 \\ \\ \frac{dy}{dx} = {x}^{2} - 4x + 3$

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