How do you find the 1st and 2nd derivative of $x^2e^x$ ?

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Answer 1 · 2016-11-14T14:23:02

$dy/dx = x(x + 2)e^x$
$(d^2y)/(dx^2) = (x^2 +4x + 2)e^x$

$dy/dx = 2x xx e^x + e^x xx x^2$

$dy/dx = 2xe^x + x^2e^x$

$dy/dx = x(x + 2)e^x$

$(d^2y)/(dx^2) = 2 xx e^x + 2x xx e^x + 2x xx e^x + x^2xxe^x$

$(d^2y)/(dx^2) = 2e^x + 2xe^x + 2xe^x + x^2e^x$

$(d^2y)/(dx^2) = 2e^x + 4xe^x + x^2e^x$

$(d^2y)/(dx^2) = (x^2 + 4x +2)e^x$

Hopefully this helps!

How do you find the 1st and 2nd derivative of x^2e^xx^2e^x?