Calculus | Derivatives of Inverse and Transcendental Functions

Overview

Some of the most important functions in calculus are inverse and transcendental functions, like logarithmic, exponential, and inverse trig functions. Understanding their derivatives is critical for solving problems involving growth, decay, and inverse relationships.

Key Concepts and Structures

Inverse Functions: If y = f^{-1}(x), then \frac{dy}{dx} = \frac{1}{f'(f^{-1}(x))}
Derivative of \ln x: \frac{d}{dx}(\ln x) = 1/x (for x > 0)
Derivative of e^x: \frac{d}{dx}(e^x) = e^x
Derivative of \log_b x: \frac{1}{x \ln b}
Inverse Trig Derivatives:
- \frac{d}{dx}(\sin^{-1} x) = \frac{1}{\sqrt{1 - x^2}}
- \frac{d}{dx}(\cos^{-1} x) = \frac{-1}{\sqrt{1 - x^2}}
- \frac{d}{dx}(\tan^{-1} x) = \frac{1}{1 + x^2}
Remember: these functions often require the chain rule when composed with other expressions.

Quick Tip

For inverse trig functions, memorize the three key derivatives and know which one applies based on the function. Be especially careful with square roots in the denominator.

Practice Problems

Differentiate: f(x) = \ln(3x^2)

Show Solution

Step 1: Use the chain rule: outer function is \ln(u), inner is 3x^2

\frac{d}{dx}(\ln(3x^2)) = \frac{1}{3x^2} \cdot 6x

Simplify: \frac{6x}{3x^2} = \frac{2}{x}

Final Answer: f'(x) = 2/x
Find the derivative of: f(x) = \tan^{-1}(2x)

Show Solution

Step 1: Use the derivative rule for inverse tangent: \frac{d}{dx}(\tan^{-1} u) = \frac{1}{1 + u^2} \cdot u'

u = 2x, so u' = 2

Final Answer: f'(x) = \frac{2}{1 + 4x^2}