Math Problem Statement

pochodna

pochodna

Solution

To find the derivative of the function f(x)=lnx+1xlogxf(x) = \sqrt{\ln x} + \frac{1}{x} \log x, we will use the rules of differentiation. Here is the step-by-step solution:

Function:

f(x)=lnx+1xlogxf(x) = \sqrt{\ln x} + \frac{1}{x} \log x

Step 1: Derivative of lnx\sqrt{\ln x}

Using the chain rule: ddxlnx=12lnxddx(lnx)=12lnx1x=12xlnx\frac{d}{dx} \sqrt{\ln x} = \frac{1}{2\sqrt{\ln x}} \cdot \frac{d}{dx} (\ln x) = \frac{1}{2\sqrt{\ln x}} \cdot \frac{1}{x} = \frac{1}{2x\sqrt{\ln x}}

Step 2: Derivative of 1xlogx\frac{1}{x} \log x

Using the product rule: ddx(1xlogx)=ddx(logxx)\frac{d}{dx} \left( \frac{1}{x} \log x \right) = \frac{d}{dx} \left( \frac{\log x}{x} \right) Let u=logxu = \log x and v=1xv = \frac{1}{x}. Then: ddx(logxx)=uv+uv\frac{d}{dx} \left( \frac{\log x}{x} \right) = u'v + uv' where u=1xu' = \frac{1}{x} and v=1x2v' = -\frac{1}{x^2}. Substituting these: ddx(logxx)=1x1x+(logx)(1x2)\frac{d}{dx} \left( \frac{\log x}{x} \right) = \frac{1}{x} \cdot \frac{1}{x} + (\log x) \cdot \left( -\frac{1}{x^2} \right) ddx(logxx)=1x2logxx2=1logxx2\frac{d}{dx} \left( \frac{\log x}{x} \right) = \frac{1}{x^2} - \frac{\log x}{x^2} = \frac{1 - \log x}{x^2}

Final Answer:

Combining both parts: f(x)=12xlnx+1logxx2f'(x) = \frac{1}{2x\sqrt{\ln x}} + \frac{1 - \log x}{x^2}

Let me know if you want further clarification or detailed steps for a specific part!


Here are 5 questions to further your understanding:

  1. Can you derive ddxlnx\frac{d}{dx} \sqrt{\ln x} independently using the chain rule?
  2. How does the product rule simplify differentiation in logxx\frac{\log x}{x}?
  3. Why does logx\log x have a derivative of 1x\frac{1}{x}?
  4. What happens to the derivative when xx \to \infty?
  5. How does the presence of lnx\ln x and logx\log x in the function affect its rate of change?

Tip: Always check the domain of the function before differentiating, especially

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Math Problem Analysis

Mathematical Concepts

Differentiation
Logarithmic Functions
Chain Rule
Product Rule

Formulas

d/dx [√(ln x)] = 1 / (2x√(ln x))
d/dx [(1/x) log x] = (1 - log x) / x^2

Theorems

Chain Rule
Product Rule

Suitable Grade Level

Undergraduate Calculus