If y = 3x^5 + 4x^4 + 2x + 3,then | vedclass

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(C) Given the function $y = 3x^5 + 4x^4 + 2x + 3$.The first derivative is $y_1 = \frac{dy}{dx} = 15x^4 + 16x^3 + 2$.The second derivative is $y_2 = \f

Vedclass · Accepted Answer

$y_6 = 0$

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(C) Given the function $y = 3x^5 + 4x^4 + 2x + 3$.The first derivative is $y_1 = \frac{dy}{dx} = 15x^4 + 16x^3 + 2$.The second derivative is $y_2 = \frac{d^2y}{dx^2} = 60x^3 + 48x^2$.The third derivative is $y_3 = \frac{d^3y}{dx^3} = 180x^2 + 96x$.The fourth derivative is $y_4 = \frac{d^4y}{dx^4} = 360x + 96$.The fifth derivative is $y_5 = \frac{d^5y}{dx^5} = 360$.The sixth derivative is $y_6 = \frac{d^6y}{dx^6} = 0$.Since the degree of the polynomial is $5$,the $n^{th}$ derivative for any $n > 5$ will be $0$. Therefore,$y_6 = 0$.

If $y = 3x^5 + 4x^4 + 2x + 3$,then

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