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

##### 1 Answer

#### 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