if f(x)=(x^2)(e^x) prove by induction that the nth derivative is (x^2 +2nx +n(n-1)(e^x)
thanks
step 1.
test if it is true for n=1
1st deriv
= x^2e^x + 2xe^x
= e^x(x^2 + 2x)
by your formula is should be
e^x(x^2 + 2(1)x + 1(0))
= e^x(x^2 + 2x) check!
(I also tested it for n=2 and 3, and it checked
step 2. assume true for n=k
i.e.
kth deriv
= e^x(x^2 + 2kx + k(k-1))
step 3.
then (k+1)th deriv should be
e^x(x^2 + 2(k+1)x + (k+1)(k))
so let's differentiate the kth deriv
(k+1)th deriv
= e^x(2x + 2k) + (x^2 + 2kx + k(k-1))e^x
= e^x[x^2 + 2kx + 2x + k(k-1) + 2k]
= e^x[x^2 + 2x(k+1) + k^2 + k]
= e^x[x^2 + 2(k+1)x + k(k+1)]
as required
Q.E.D.
To prove by induction that the nth derivative of f(x) = x^2 * e^x is (x^2 + 2nx + n(n-1) * e^x), we will follow these steps:
Step 1: Base Case
First, we need to prove the statement for the base case (n = 1).
The first derivative of f(x) with respect to x is given by:
f'(x) = d/dx(x^2 * e^x)
Using the product rule, we can differentiate f(x) with respect to x:
f'(x) = 2x * e^x + x^2 * e^x
This matches (x^2 + 2x + 1*(1-1) * e^x) as n = 1, so the base case is true.
Step 2: Inductive Hypothesis
Assume the statement is true for the nth derivative, i.e., the nth derivative of f(x) is given by:
f^n(x) = (x^2 + 2nx + n(n-1) * e^x) ----(1)
Step 3: Inductive Step
We need to prove that the statement is true for (n+1)th derivative, i.e., the (n+1)th derivative of f(x) is given by:
f^(n+1)(x) = (x^2 + 2(n+1)x + (n+1)(n)((n+1)-1) * e^x) ----(2)
To do this, we differentiate Equation (1) with respect to x:
f^n(x) = (x^2 + 2nx + n(n-1) * e^x)
Differentiating both sides with respect to x gives:
f^(n+1)(x) = d/dx[(x^2 + 2nx + n(n-1) * e^x)]
Differentiating each term using the product rule and simplifying the expression, we get:
f^(n+1)(x) = (2x + 2n + 2n(n-1) + n(n-1)) * e^x + (x^2 + 2nx + n(n-1)) * e^x
f^(n+1)(x) = (x^2 + 2(n+1)x + (n+1)(n)((n+1)-1) * e^x)
This matches Equation (2), and thus the statement is true for (n+1).
Hence, the statement is proven by induction.
To prove this statement by induction, we need to first establish the base case and then prove the inductive step.
Step 1: Base Case
Let's start by proving the statement for n = 1, which is the base case.
Given f(x) = (x^2)(e^x), we need to find the first derivative, f'(x).
f'(x) = d/dx[(x^2)(e^x)]
Using the product rule, we can differentiate this function:
f'(x) = (d/dx)(x^2)(e^x) + (x^2)(d/dx)(e^x)
Simplifying this expression:
f'(x) = 2x(e^x) + (x^2)(e^x)
This matches the expression for n = 1: (x^2 + 2x + e^x). Thus, the base case holds true.
Step 2: Inductive Step
Now, let's assume that the statement is true for n = k, i.e., the kth derivative of f(x) is given by:
f^(k)(x) = (x^2 + 2kx + k(k-1))(e^x)
We need to prove that the statement holds true for n = k+1, i.e., we need to show that the (k+1)th derivative of f(x) is given by:
f^(k+1)(x) = (x^2 + 2(k+1)x + (k+1)(k)(e^x)
To do this, we'll differentiate the assumed expression for n = k.
f^(k+1)(x) = (d/dx)[(x^2 + 2kx + k(k-1))(e^x)]
Using the product rule and the assumption, we can simplify this expression:
f^(k+1)(x) = 2k(e^x) + (x^2 + 2kx + k(k-1))(e^x) + (x^2 + 2kx + k(k-1))(e^x)
Simplifying further:
f^(k+1)(x) = (2k + 2)(e^x) + (2k + 1)(x^2 + 2kx + k(k-1))(e^x)
Expanding the quadratic term:
f^(k+1)(x) = (2k + 2)(e^x) + (2k + 1)(x^2 + 2kx + k^2 - k)(e^x)
Rearranging and combining like terms:
f^(k+1)(x) = (x^2 + 2(k+1)x + (k+1)k)(e^x)
This matches the expression for n = k+1. Therefore, the statement holds for n = k+1.
Step 3: Conclusion
Since we have proven that the statement is true for the base case (n = 1) and have shown that if it holds for n = k, it also holds for n = k+1, we can conclude that the statement is true for all positive integers n by mathematical induction.