What is the n^(th) derivative of x^2e^x ?

1 Answer
Aug 11, 2017

(d^(n))/(dx^n) x^2e^x = y^((n)) (x)
" " = (x^2+2nx+n(n-1))e^x

Explanation:

We have:

y(x)=x^2e^x ..... [A]

Differentiating wrt x, and applying the product rule we get:

y' \ \ \ \ \= x^2e^x+2xe^x
\ \ \ \ \ \ \ \ = (x^2+2x)e^x ..... [B]

Differentiating again wrt x, and applying the product rule we get:

y'' \ \ \ \= (x^2+2x)e^x + (2x+2)e^x
\ \ \ \ \ \ \ \ = (x^2+4x+2)e^x

Differentiating again wrt x, and applying the product rule we get:

y''' \ = (x^2+2x)e^x + (2x+2)e^x
\ \ \ \ \ \ \ \ = (x^2+4x+2)e^x + (2x+4)e^x
\ \ \ \ \ \ \ \ = (x^2+6x+6)e^x

Differentiating again wrt x, and applying the product rule we get:

y^((4)) = (x^2+2x)e^x + (2x+2)e^x
\ \ \ \ \ \ \ = (x^2+6x+6)e^x + (2x+6)e^x
\ \ \ \ \ \ \ = (x^2+8x+12)e^x

Prediction

Based on the above results, it "appears" as if the n^(th) derivative is given by:

y^((n)) = (x^2+2nx+n^2-n)e^x

So let us test this prediction using Mathematical Induction:

Induction Proof - Hypothesis

We aim to prove by Mathematical Induction that for n in NN that:

(d^(n))/(dx^n) x^2e^x = y^((n)) = (x^2+2nx+n^2-n)e^x
" " = (x^2+2nx+n(n-1))e^x ..... [C]

Induction Proof - Base case:

We will show that the given result holds for:

n=0 : (the function)
n=1 : (the first derivative)

When n=0 the given result gives:

y^((0)) = (x^2)e^x = y(x) (from [A])

When n=1 the given result gives:

y^((1)) = (x^2+2x)e^x = y'(x) (from [B])

So the given result is true when n=1 (and in fact n=0)

Induction Proof - General Case

Now, Let us assume that the given result [C] is true when n=m, for some m in NN, m gt 1, in which case for this particular value of m we have:

(d^(m))/(dx^m) x^2e^x = y^((m)) = (x^2+2mx+m^2-m)e^x

So, differentiating the above, using the product rule, we get:

(d^(m+1))/(dx^(m+1)) x^2e^x = (x^2+2mx+m^2-m)e^x + (2x+2m)e^x
" " = (x^2+2mx+2x+m^2-m+2m)e^x
" " = (x^2+2((m+1)x)+m^2+m)e^x
" " = (x^2+2((m+1)x)+(m+1)(m))e^x

Which is the given result [C] with n=m+1

Induction Proof - Summary

So, we have shown that if the given [C] result is true for n=m, then it is also true for n=m+1. But we initially showed that the given result was true for n=0 and n=1 so it must also be true for n=2, n=3, n=4, ... and so on.

Hence, by the process of mathematical induction the given result is true for n in NN QED