If y = x^2 e^{mx},where m is a constant,then | vedclass

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(A) Given $y = x^2 e^{mx}$.Differentiating with respect to $x$ using the product rule:$\frac{dy}{dx} = 2x e^{mx} + x^2 (m e^{mx}) = e^{mx} (m x^2 + 2x

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$m e^{mx} (m^2 x^2 + 6mx + 6)$

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(A) Given $y = x^2 e^{mx}$.Differentiating with respect to $x$ using the product rule:$\frac{dy}{dx} = 2x e^{mx} + x^2 (m e^{mx}) = e^{mx} (m x^2 + 2x)$.Differentiating again with respect to $x$:$\frac{d^2y}{dx^2} = m e^{mx} (m x^2 + 2x) + e^{mx} (2mx + 2) = e^{mx} (m^2 x^2 + 2mx + 2mx + 2) = e^{mx} (m^2 x^2 + 4mx + 2)$.Differentiating a third time with respect to $x$:$\frac{d^3y}{dx^3} = m e^{mx} (m^2 x^2 + 4mx + 2) + e^{mx} (2m^2 x + 4m) = e^{mx} (m^3 x^2 + 4m^2 x + 2m + 2m^2 x + 4m) = e^{mx} (m^3 x^2 + 6m^2 x + 6m)$.Factoring out $m$:$\frac{d^3y}{dx^3} = m e^{mx} (m^2 x^2 + 6mx + 6)$.

If $y = x^2 e^{mx}$,where $m$ is a constant,then $\frac{d^3y}{dx^3} = $

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