Find f '(x) and f ''(x).
f(x) = x^(9\/ 2) e^x
To find the derivative of f(x) = x^(9/2) e^x, we can use the product rule and the chain rule.
The product rule states that if we have two functions u(x) and v(x), the derivative of their product is given by (u * v)' = u' * v + u * v'.
Let's differentiate each term separately:
For the first term, x^(9/2), we can apply the power rule which states that the derivative of x^n is given by n * x^(n-1). In this case, n = 9/2, so the derivative of x^(9/2) is (9/2) * x^(9/2 - 1) = (9/2) * x^(7/2).
For the second term, e^x, the derivative is simply e^x since the derivative of e^x is itself.
Now, using the product rule, we have:
f'(x) = (9/2) * x^(7/2) * e^x + x^(9/2) * e^x
To find the second derivative, f''(x), we differentiate f'(x):
For the first term, (9/2) * x^(7/2) * e^x, we again apply the power rule to x^(7/2) and leave e^x unchanged. The derivative is then (9/2) * (7/2) * x^(7/2 - 1) * e^x = (63/4) * x^(5/2) * e^x.
For the second term, x^(9/2) * e^x, we use the product rule again. The derivative is x^(7/2) * e^x + x^(9/2) * e^x.
Combining these terms, we have:
f''(x) = (63/4) * x^(5/2) * e^x + x^(7/2) * e^x + x^(9/2) * e^x
Therefore, f'(x) = (9/2) * x^(7/2) * e^x + x^(9/2) * e^x, and f''(x) = (63/4) * x^(5/2) * e^x + x^(7/2) * e^x + x^(9/2) * e^x.