Derivatives of Exponential and Logarithm Functions — Question 5

Question 5

Let the function $f(x) = \ln\left(\frac{x^2 \sqrt{e^x + 1}}{\ln x}\right)$, for $x > 1$.

• Simplify the expression using logarithmic properties.

• Find the derivative $f'(x)$.

Question 5 - Solution

We are given: $$f(x) = \ln\left(\frac{x^2 \sqrt{e^x + 1}}{\ln x}\right), \quad x > 1$$

(a) Simplify the expression:

Use logarithmic identities: $$\ln\left(\frac{A \cdot B}{C}\right) = \ln(A) + \ln(B) - \ln(C)$$

Apply to this expression: $$f(x) = \ln(x^2) + \ln\left((e^x + 1)^{1/2}\right) - \ln(\ln x)$$

Now simplify powers: $$f(x) = 2\ln x + \frac{1}{2}\ln(e^x + 1) - \ln(\ln x)$$

(b) Differentiate $f(x)$:

Differentiate term-by-term:

$\frac{d}{dx}[2\ln x] = \frac{2}{x}$

$\frac{d}{dx}\left[\frac{1}{2} \ln(e^x + 1)\right] = \frac{1}{2} \cdot \frac{e^x}{e^x + 1}$

$\frac{d}{dx}[\ln(\ln x)] = \frac{1}{x \ln x}$

Putting it all together: $$f'(x) = \frac{2}{x} + \frac{1}{2} \cdot \frac{e^x}{e^x + 1} - \frac{1}{x \ln x}$$

Final Answer: $$\boxed{ f'(x) = \frac{2}{x} + \frac{e^x}{2(e^x + 1)} - \frac{1}{x \ln x} }$$