What is the #n^(th)# derivative of #x^2e^x #?
1 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
# y' \ \ \ \ \= x^2e^x+2xe^x #
# \ \ \ \ \ \ \ \ = (x^2+2x)e^x # ..... [B]
Differentiating again wrt
# y'' \ \ \ \= (x^2+2x)e^x + (2x+2)e^x #
# \ \ \ \ \ \ \ \ = (x^2+4x+2)e^x #
Differentiating again wrt
# 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
# 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
# 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
# (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
# y^((0)) = (x^2)e^x = y(x) # (from [A])
When
# y^((1)) = (x^2+2x)e^x = y'(x) # (from [B])
So the given result is true when
Induction Proof - General Case
Now, Let us assume that the given result [C] is true when
# (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
Induction Proof - Summary
So, we have shown that if the given [C] result is true for
Hence, by the process of mathematical induction the given result is true for