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

1 Answer
Aug 11, 2017

Answer:

# (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