Compute f'(e) where f(x)=3^(xlnx)
f'(x) = ln3(x(1/x) + lnx)(3^(xlnx))
= ln3(1 + lnx)(3^(xlnx))
so f'(e) = ln3(1+lne)(3^(elne))
= ln3(2)(3^(e))
To compute f'(e) for the function f(x) = 3^(xlnx), we need to find the derivative of f(x) and then evaluate it at x = e.
Let's start by taking the derivative of f(x) using the chain rule:
Step 1: Rewrite f(x) = 3^(xlnx) as f(x) = e^[ln(3^(xlnx))].
- Here, we use the property a^(bc) = (a^b)^c to rewrite 3^(xlnx) as e^(ln(3^(xlnx))).
Step 2: Apply the chain rule.
- The chain rule states that if we have a composite function f(g(x)), then its derivative is given by f'(g(x)) * g'(x).
- In this case, we have f(g(x)) = e^[ln(3^(xlnx))], so we need to find f'(g(x)) and g'(x).
Step 3: Find f'(g(x)).
- Let h(x) = ln(3^(xlnx)).
- To differentiate h(x), we use the chain rule again:
h'(x) = (3^(xlnx)) * (ln(3) * d/dx(xlnx)).
- Now, we need to find d/dx(xlnx):
Using the product rule, d/dx(xlnx) = 1 * ln(x) + x * (1/x) = ln(x) + 1.
- Therefore, h'(x) = (3^(xlnx)) * (ln(3) * (ln(x) + 1)).
- Since f(g(x)) = e^[ln(3^(xlnx))], we have f'(g(x)) = e^[ln(3^(xlnx))] * ln(3) * (ln(x) + 1).
Step 4: Find g'(x).
- g(x) = xlnx, so we differentiate g(x) using the product rule:
g'(x) = ln(x) + x * (1/x) = ln(x) + 1.
Step 5: Evaluate f'(e).
- Plug in x = e into f'(g(x)) and g'(x):
f'(x) = e^[ln(3^(xlnx))] * ln(3) * (ln(x) + 1)
g'(x) = ln(x) + 1
- So, f'(e) = e^[ln(3^(eln(e)))] * ln(3) * (ln(e) + 1).
Simplifying f'(e), we have:
- f'(e) = e^[ln(3^e)] * ln(3) * (1 + 1)
- Since ln(3^e) = e * ln(3), we can rewrite it as:
f'(e) = e^[e * ln(3)] * ln(3) * 2
Therefore, f'(e) = e^[e * ln(3)] * ln(3) * 2.