# Given the function f(x)=(2x^2-3x+2)*e^x, differentiate the function n times.

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We'll find the n-th derivative of the given function, using Leibnitz formula. We can apply Leibnitz formula since the given function is a product of 2 functions:

u(x) = 2x^2-3x+2 and v(x) = e^x

According to Leibniz formula:

[u(x)*v(x)]^(n) =C(n,0)*[u(x)]^(n)*v(x) + C(n,1)*[u(x)]^(n-1)*v'(x) + C(n,2)*[u(x)]^(n-2)*v"(x) + ... + C(n,n)*u(x)*[v(x)]^(n).

We notice that [v(x)]^(k) = e^x.

We also notice that u'(x) = 4x-3

u"(x) =4

u"'(x) = 0

So, we'll get only 3 terms in the Leibniz's formula:

[f(x)]^(n) = e^x[2x^2 - 3x + 2 + n(4x-3) + 4n(n-1)/2]

[f(x)]^(n) = e^x[2x^2 + (4n+3)x + 2n^2 - 5n + 2]

**After differentiating the function f(x) n times, the n-th derivative of the function is: [f(x)]^(n) = e^x[2x^2 + (4n+3)x + 2n^2 - 5n + 2].**