Question

For the following exercises, find the derivative \(d y / d x\). (You can use a calculator to plot the function and the derivative to confirm that it is correct.) $$ \text { [T] } y=\ln (\tan x) $$

Short answer

The derivative is \( \csc(x) \).

Step-by-step solution

Understanding the Problem

We need to find the derivative \( \frac{d y}{dx} \) of the function \( y = \ln (\tan x) \). This involves using rules for differentiation such as the derivative of the natural logarithm and the chain rule.

Applying the Chain Rule

The derivative of \( \ln(u) \) where \( u = \tan(x) \) is \( \frac{1}{u} \cdot \frac{du}{dx} \). Thus, we have \( \frac{d}{dx} \left\{ \ln(\tan x) \right\} = \frac{1}{\tan(x)} \cdot \frac{d}{dx}(\tan x) \).

Differentiating \( \tan(x) \)

The derivative of \( \tan(x) \) with respect to \( x \) is \( \sec^2(x) \), so \( \frac{d}{dx}(\tan x) = \sec^2(x) \).

Combining the Results

Using the results from the previous steps, we substitute back into our chain rule formula: \( \frac{d}{dx} \left\{ \ln(\tan x) \right\} = \frac{1}{\tan(x)} \cdot \sec^2(x) \).

Simplifying the Expression

We know that \( \sec(x) = \frac{1}{\cos(x)} \) and \( \tan(x) = \frac{\sin(x)}{\cos(x)} \). Thus, \( \frac{sec^2(x)}{\tan(x)} = \frac{1/\cos^2(x)}{\sin(x)/\cos(x)} \). Simplifying, we get \( \frac{\cos(x)}{\sin(x) \cdot \cos(x)} = \frac{1}{\sin(x)} = \csc(x) \). Thus, \( \frac{d y}{d x} = \csc(x) \).

Key concepts

Chain Rule

The chain rule is a fundamental concept in calculus that allows us to find the derivative of composite functions. Composite functions are those where one function is nested inside another, like our example, where the function is the natural logarithm of the tangent function. The chain rule is particularly powerful because it enables us to handle such multi-layered functions in one go. When using the chain rule, we look for an inner function and an outer function. If we have a function of the form \( y = f(g(x)) \), where \( g(x) \) is the inner function and \( f(u) \) is the outer function with \( u = g(x) \), the derivative of \( y \) with respect to \( x \) is given by:\[ \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \]In our specific problem, \( f(u) = \ln(u) \) and \( g(x) = \tan(x) \). When we differentiate, \( f'(u) = \frac{1}{u} \) and \( g'(x) = \sec^2(x) \). We then multiply these derivatives to get the derivative of the composite function. Using the chain rule efficiently unravels complex differentiation problems.

Natural Logarithm Differentiation

The natural logarithm function, denoted as \( \ln(x) \), has a derivative that is quite straightforward but essential to grasp. The derivative of \( \ln(x) \) is \( \frac{1}{x} \). This rule is applied when differentiating any natural logarithm function, provided the argument is a differentiable function itself. In our problem, the natural logarithm applies not to \( x \) directly, but to another function, \( \tan(x) \). This is why the chain rule is necessary: it allows us to differentiate the inner function \( \tan(x) \) and then multiply it with the derivative of the outer function \( \ln(u) \). The formula for differentiating \( \ln(u) \) in terms of \( x \) is:\[ \frac{d}{dx} \left( \ln(u) \right) = \frac{1}{u} \cdot \frac{du}{dx} \]Applying this to our problem, \( u = \tan(x) \), hence the expression transforms to \( \frac{1}{\tan(x)} \cdot \sec^2(x) \). This showcases the tight interplay between logarithmic differentiation and other functions.

Trigonometric Derivatives

Trigonometric derivatives are integral to solving calculus problems involving trigonometric functions. In the exercise, we needed to differentiate \( \tan(x) \), which is a standard trigonometric function. The derivative of \( \tan(x) \) itself is \( \sec^2(x) \). This result originates from the fundamental identities of trigonometry and the derivatives of sine and cosine functions. Specifically, since \( \tan(x) = \frac{\sin(x)}{\cos(x)} \), applying the quotient rule yields \( \sec^2(x) = \left( \frac{1}{\cos(x)} \right)^2 \).In this exercise, after differentiating \( \tan(x) \) and simplifying the result, we use another trigonometric fact that relates secant and tangent functions to sine and cosine functions:\[ \sec(x) = \frac{1}{\cos(x)} \quad \text{and} \quad \tan(x) = \frac{\sin(x)}{\cos(x)} \]Thus, the derivative simplifies to \( \csc(x) \), which is \( \frac{1}{\sin(x)} \). Understanding these trigonometric derivatives is vital for solving calculus problems involving complex expressions efficiently.