Problem 16 - Calculus Single Variable

Question

Calculate the derivative with respect to \(x\) of the given expression. \(\ln (\sec (x))\)

Step-by-Step Solution

Answer

The derivative of \( \ln(\sec(x)) \) with respect to \( x \) is \( \tan(x) \).

Step 1: Recall the chain rule

The chain rule is a fundamental rule for differentiating composite functions. If you have a function that is a composition of two functions, say \( f(g(x)) \), the chain rule states that the derivative is \( f'(g(x)) \cdot g'(x) \). We'll apply this rule for differentiating \( \ln(\sec(x)) \).

Step 2: Differentiate the outer function

Identify \( f(u) = \ln(u) \) where \( u = \sec(x) \). The derivative of \( \ln(u) \) with respect to \( u \) is \( \frac{1}{u} \). So, the derivative of \( \ln(\sec(x)) \) with respect to \( \sec(x) \) is \( \frac{1}{\sec(x)} \) or \( \cos(x) \).

Step 3: Differentiate the inner function

The next step is to differentiate \( \sec(x) \) with respect to \( x \). Recall that the derivative of \( \sec(x) \) is \( \sec(x)\tan(x) \).

Step 4: Apply the chain rule

Using the chain rule, multiply the derivatives from Step 2 and Step 3 together: \( \cos(x) \cdot \sec(x)\tan(x) \). This simplifies to \( \tan(x) \) since \( \cos(x) \cdot \sec(x) = 1 \).

Step 5: Simplify

The final expression for the derivative of \( \ln(\sec(x)) \) with respect to \( x \) is \( \tan(x) \).

Key Concepts

Chain Rule

When dealing with derivatives of composite functions, the chain rule becomes your go-to tool. Consider having a function nested within another, like in our exercise where we see \(\ln(\sec(x))\).The chain rule helps you find the derivative of such intricate expressions. It can be understood as:

• Differentiate the outer function leaving the inner function untouched.
• Multiply by the derivative of the inner function.

Breaking it down for our problem, identify \( f(u) = \ln(u) \) just like in our exercise. The outer function here is \( \ln(u) \) and the inner function is \( \sec(x) \). Then proceed with each separately. The overall derivative is a multiplication of these results.

Derivative of Trigonometric Functions

Trigonometric functions often pop up in calculus, and knowing their derivatives by heart can ease many calculations.Let's zoom in on \( \sec(x) \), which is the reciprocal of \( \cos(x) \). Its derivative is particularly useful: \[ \frac{d}{dx}[\sec(x)] = \sec(x)\tan(x) \]Once you remember this, finding derivatives involving \( \sec(x) \) becomes straightforward. In our exercise, after differentiating the outer \( \ln \) function, you move on to \( \sec(x)\).Understanding these core results allows you to evaluate complex derivatives with confidence and accuracy.

Logarithmic Differentiation

Logarithmic differentiation can be tremendously helpful when working with complex products, quotients, or powers. Even though it's most popular for tricky algebra concerns, our problem shows its role in simplifying expressions. At the heart, it's maximizing algebraic structure to simplify derivatives.Whenever you differentiate \( \ln(u) \), understand that:

• The derivative of \( \ln(u) \) with respect to \( u \) is \( \frac{1}{u} \).

So for \( \ln(\sec(x)) \), the transformation becomes straightforward when following the chain rule. Once derivatives of trigonometric elements come in, you multiply, apply, and simplify.