Problem 17 - Essential Calculus Early Transcendentals

Question

Differentiate the function. $$ y=\tan [\ln (a x+b)] $$

Step-by-Step Solution

Answer

The derivative is \( \frac{a \sec^2(\ln(ax + b))}{ax + b} \).

Step 1: Identify the Function and Its Composition

The given function is \( y = \tan [\ln (ax + b)] \). This is a composition of functions where the outer function is \( \tan(u) \) and the inner function is \( u = \ln(ax + b) \).

Step 2: Differentiate the Outer Function

Differentiate the outer function \( \tan(u) \) with respect to \( u \). The derivative is \( \sec^2(u) \). So, \( \frac{d}{du}\tan(u) = \sec^2(u) \).

Step 3: Differentiate the Inner Function

Differentiate the inner function \( u = \ln(ax + b) \) with respect to \( x \). Using the chain rule for differentiation, \( \frac{d}{dx}\ln(ax + b) = \frac{1}{ax + b} \cdot a = \frac{a}{ax + b} \).

Step 4: Apply the Chain Rule

Using the chain rule \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \), combine the derivatives from Step 2 and Step 3. We have \( \frac{dy}{dx} = \sec^2(\ln(ax + b)) \cdot \frac{a}{ax + b} \).

Step 5: Final Expression for the Derivative

Combining everything, the derivative of \( y = \tan[\ln(ax + b)] \) with respect to \( x \) is \( \frac{dy}{dx} = \frac{a \sec^2(\ln(ax + b))}{ax + b} \).

Key Concepts

Chain Rule

The chain rule is a fundamental concept in calculus. It's used for differentiating composite functions. A composite function is a function within another function, like our exercise's function, \( y = \tan[\ln(ax+b)] \).

To use the chain rule, identify the inner and outer functions. In our example:

• The inner function is \( u = \ln(ax + b) \).
• The outer function is \( \tan(u) \).

The chain rule formula is \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \), where:

• \( \frac{dy}{du} \) is the derivative of the outer function with respect to the inner function.
• \( \frac{du}{dx} \) is the derivative of the inner function with respect to \( x \).

This method allows us to effectively find the derivative of complex compositions.

Apply the chain rule whenever you differentiate functions wrapped within each other. This powerful tool simplifies these processes tremendously.

Trigonometric Functions

Trigonometric functions, such as sine, cosine, and tangent, are fundamental in calculus and various branches of mathematics. Differentiating these functions is essential for solving many calculus problems.

In our exercise, the focus is on the tangent function, \( \tan(u) \). To differentiate \( \tan(u) \), we use the identity:

• \( \frac{d}{du}\tan(u) = \sec^2(u) \).

This derivative arises from the identity \( \sec(u) \), which is the reciprocal of the cosine function.

Knowing the derivatives of trigonometric functions can greatly assist you in solving calculus problems that involve these functions. Whether you are dealing with simple equations or complex compositions, trigonometric functions are key elements in many solutions.

Logarithmic Differentiation

Logarithmic differentiation is a method used for differentiating functions involving logarithms, especially when they are part of a larger composition.

In our example, the inner function \( \ln(ax + b) \) requires its own differentiation:

• The derivative of \( \ln(ax + b) \) with respect to \( x \) is \( \frac{1}{ax + b} \cdot a = \frac{a}{ax + b} \).

This result comes from the property that the derivative of \( \ln(x) \) is \( \frac{1}{x} \).

Logarithmic differentiation can simplify the differentiation process for products and quotients or functions raised to variable exponents, but even in nested functions, it proves useful. Grasping this concept helps in navigating through intricate calculus problems efficiently.