Question

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

Answer 1

`(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