if f(x)=(x^2)(e^x) prove by induction that the nth derivative is (x^2 +2nx +n(n-1)(e^x)
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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:
1. Base case: We'll start by proving the statement for n = 0. In this case, the 0th derivative is the original function itself, so f(x) = (x^2)(e^x). Therefore, the base case holds true.
2. Inductive hypothesis: Assume that the statement is true for some integer k ≥ 0. That is, assume that the kth derivative of f(x) is given by (x^2 + 2kx + k(k-1))(e^x).
3. Inductive step: We'll now prove that if the statement holds for k, it also holds for k+1.
We know that the (k+1)th derivative of f(x) is the derivative of the kth derivative:
f^(k+1)(x) = [d/dx][(x^2 + 2kx + k(k-1))(e^x)]
To differentiate this expression, we'll use the product rule.
a. Differentiate (x^2 + 2kx + k(k-1)):
The derivative of (x^2 + 2kx + k(k-1)) with respect to x is 2x + 2k.
b. Multiply by (e^x):
To differentiate the product, we use the product rule again. We differentiate (e^x) with respect to x, which is simply (e^x).
So, we have:
f^(k+1)(x) = [(2x + 2k)(e^x)] + [(x^2 + 2kx + k(k-1))(e^x)]
Expanding and simplifying, we get:
f^(k+1)(x) = (2x + 2k)e^x + (x^2 + 2kx + k(k-1))(e^x)
= (2x + 2k + x^2 + 2kx + k(k-1))(e^x)
= (x^2 + 2(k+1)x + (k+1)k)(e^x)
Therefore, the (k+1)th derivative of f(x) is given by (x^2 + 2(k+1)x + (k+1)k)(e^x).
4. Conclusion: By induction, we have proven that for all integers n ≥ 0, the nth derivative of f(x) = (x^2)(e^x) is (x^2 + 2nx + n(n-1))(e^x).