Math Problem Statement

Let f(x) = x^√x. Find f'(9).

Solution

The given function is f(x)=xxf(x) = x^{\sqrt{x}}, and you are tasked with finding f(9)f'(9), the derivative of the function evaluated at x=9x = 9.

Step 1: Rewrite the function

We can express f(x)=xxf(x) = x^{\sqrt{x}} in a more manageable form for differentiation by using logarithms. First, recall that: xx=exln(x)x^{\sqrt{x}} = e^{\sqrt{x} \ln(x)} Thus, f(x)=exln(x)f(x) = e^{\sqrt{x} \ln(x)}.

Step 2: Differentiate the function using the chain rule

We will now differentiate f(x)=exln(x)f(x) = e^{\sqrt{x} \ln(x)}. Let g(x)=xln(x)g(x) = \sqrt{x} \ln(x), so: f(x)=eg(x)f(x) = e^{g(x)}

To find f(x)f'(x), apply the chain rule: f(x)=eg(x)g(x)f'(x) = e^{g(x)} \cdot g'(x)

Step 3: Differentiate g(x)=xln(x)g(x) = \sqrt{x} \ln(x)

We need to compute g(x)g'(x). Use the product rule since g(x)g(x) is the product of x\sqrt{x} and ln(x)\ln(x). Recall that x=x1/2\sqrt{x} = x^{1/2}, so we have: g(x)=x1/2ln(x)g(x) = x^{1/2} \ln(x)

Using the product rule: g(x)=12x1/2ln(x)+x1/21x=ln(x)2x+1xg'(x) = \frac{1}{2}x^{-1/2} \ln(x) + x^{1/2} \cdot \frac{1}{x} = \frac{\ln(x)}{2\sqrt{x}} + \frac{1}{\sqrt{x}}

Step 4: Combine results

Now, we can express the derivative of f(x)f(x) as: f(x)=exln(x)(ln(x)2x+1x)f'(x) = e^{\sqrt{x} \ln(x)} \left( \frac{\ln(x)}{2\sqrt{x}} + \frac{1}{\sqrt{x}} \right) Since f(x)=exln(x)=xxf(x) = e^{\sqrt{x} \ln(x)} = x^{\sqrt{x}}, we can write: f(x)=xx(ln(x)2x+1x)f'(x) = x^{\sqrt{x}} \left( \frac{\ln(x)}{2\sqrt{x}} + \frac{1}{\sqrt{x}} \right)

Step 5: Evaluate at x=9x = 9

Now substitute x=9x = 9 into the derivative:

  • ln(9)=2ln(3)\ln(9) = 2 \ln(3)
  • 9=3\sqrt{9} = 3
  • 99=93=7299^{\sqrt{9}} = 9^3 = 729

Thus: f(9)=729(ln(9)23+13)=729(2ln(3)6+13)f'(9) = 729 \left( \frac{\ln(9)}{2 \cdot 3} + \frac{1}{3} \right) = 729 \left( \frac{2 \ln(3)}{6} + \frac{1}{3} \right)

This simplifies further to: f(9)=729(ln(3)3+13)=729ln(3)+13f'(9) = 729 \left( \frac{\ln(3)}{3} + \frac{1}{3} \right) = 729 \cdot \frac{\ln(3) + 1}{3}

Finally: f(9)=243(ln(3)+1)f'(9) = 243 (\ln(3) + 1)

Thus, the derivative at x=9x = 9 is: f(9)=243(ln(3)+1)f'(9) = 243 (\ln(3) + 1)

Let me know if you would like further clarifications or details!

Here are five related questions:

  1. How do we simplify expressions involving logarithms and powers?
  2. What is the product rule for derivatives and how is it applied?
  3. Can you differentiate functions with logarithmic and power terms using implicit differentiation?
  4. What are some common derivative rules for exponential functions?
  5. How would you evaluate logarithmic functions for specific values?

Tip: Always break down complex expressions into simpler components before differentiating, especially when logarithms or powers are involved.

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Math Problem Analysis

Mathematical Concepts

Exponential Functions
Logarithmic Differentiation
Chain Rule
Product Rule

Formulas

f(x) = e^{√x * ln(x)}
Product Rule: (u * v)' = u'v + uv'
Chain Rule: (f(g(x)))' = f'(g(x)) * g'(x)

Theorems

Logarithmic Differentiation Theorem
Chain Rule for Exponentials

Suitable Grade Level

College-Level Calculus or Advanced High School Calculus