Answer:
\(\displaystyle \frac{dy}{dx} = \frac{1}{x \Big( (\ln x)^2 + 1 \Big)}\)
General Formulas and Concepts:
Calculus
Differentiation
- Derivatives
- Derivative Notation
Derivative Rule [Chain Rule]: \(\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)\)
Step-by-step explanation:
Step 1: Define
Identify
\(\displaystyle y = \arctan (\ln x)\)
Step 2: Differentiate
- Trigonometric Differentiation [Derivative Rule - Chain Rule]: \(\displaystyle y' = \frac{(\ln x)'}{(\ln x)^2 + 1}\)
- Logarithmic Differentiation: \(\displaystyle y' = \frac{\frac{1}{x}}{(\ln x)^2 + 1}\)
- Simplify: \(\displaystyle y' = \frac{1}{x \Big( (\ln x)^2 + 1 \Big)}\)
Topic: AP Calculus AB/BC (Calculus I/I + II)
Unit: Differentiation
