Problem 53 - Thomas Calculus

Question

In Exercises \(49-70\) , find the derivative of \(y\) with respect to the appropriate variable. $$ y=\sec ^{-1}(2 s+1) $$

Step-by-Step Solution

Answer

The derivative of \(y = \sec^{-1}(2s+1)\) is \(\frac{dy}{ds} = \frac{1}{|2s+1|\sqrt{s(s+1)}}\).

Step 1: Identify the Formula for Derivative of Inverse Functions

To differentiate the inverse secant function, use the formula: \[\frac{d}{dx} \left( \sec^{-1}(u) \right) = \frac{1}{|u| \sqrt{u^2 - 1}} \cdot \frac{du}{dx}\]where \(u\) is a function of \(x\). Here \(u = 2s + 1\).

Step 2: Differentiate the Inner Function

Find \(\frac{du}{ds}\):\[\frac{d}{ds}(2s + 1) = 2\]This is the derivative of the inner function \(2s + 1\) with respect to \(s\).

Step 3: Substitute into the Derivative Formula

Substitute \(u = 2s + 1\) and \(\frac{du}{ds} = 2\) into the derivative formula:\[\frac{dy}{ds} = \frac{1}{|2s + 1| \sqrt{(2s + 1)^2 - 1}} \cdot 2\]

Step 4: Simplify the Expression

Simplify the expression under the square root:\[(2s + 1)^2 - 1 = 4s^2 + 4s + 1 - 1 = 4s^2 + 4s\]Thus, the derivative simplifies to:\[\frac{dy}{ds} = \frac{2}{|2s + 1| \sqrt{4s^2 + 4s}}\]

Step 5: Factorization and Final Simplification

Factor the expression under the square root:\[\sqrt{4s^2 + 4s} = \sqrt{4s(s + 1)} = 2\sqrt{s(s + 1)}\]Substitute back:\[\frac{dy}{ds} = \frac{2}{|2s + 1| \cdot 2 \sqrt{s(s + 1)}} = \frac{1}{|2s + 1| \sqrt{s(s + 1)}}\]

Key Concepts

Inverse Functions

Inverse functions are a key concept in calculus, allowing us to reverse the effect of a function. Simply put, if a function maps an input to an output, its inverse reverses this process. Suppose we have a function \( f(x) \). The inverse function, denoted \( f^{-1}(x) \), satisfies the condition \( f(f^{-1}(x)) = x \) and \( f^{-1}(f(x)) = x \). Understanding inverse functions is crucial when differentiating them, as it involves understanding how inputs and outputs switch roles. Calculus often requires finding the derivative of inverse functions to understand how changes in outputs affect changes in inputs. Calculators or tables don't always have these derivatives, so knowing common inverse derivatives like \( \sec^{-1}(x) \), as we'll explore, is essential.

Derivative of Inverse Secant

To differentiate inverse trigonometric functions like inverse secant, we use specific derivative formulas tailored to these functions. For the inverse secant function \( \sec^{-1}(u) \), the derivative formula is:\[ \frac{d}{dx} \left( \sec^{-1}(u) \right) = \frac{1}{|u| \sqrt{u^2 - 1}} \cdot \frac{du}{dx} \]This formula accounts for how changes in \( u \) affect the value of \( y = \sec^{-1}(u) \). The expression \(|u| \sqrt{u^2 - 1}\) ensures the function's validity for all permissible values of \( u \), avoiding undefined scenarios caused by division by zero or negative square roots. Always check the domain and ensure that \( u^2 \ge 1 \) so that the function is defined.

Chain Rule for Differentiation

The chain rule is a fundamental tool for differentiating composite functions. It states that the derivative of a composite function \( f(g(x)) \) is the derivative of \( f \) with respect to \( g \) times the derivative of \( g \) with respect to \( x \). Mathematically, it is expressed as:\[ \frac{d}{dx} f(g(x)) = f'(g(x)) \cdot g'(x) \]In our problem, \( y = \sec^{-1}(2s+1) \), the outer function is \( \sec^{-1}(u) \) and the inner function is \( u = 2s + 1 \). Applying the chain rule involves calculating the derivative of the inner function \( g'(x) \), which gives \( \frac{du}{ds} = 2 \), and substituting it into the derivative of the outer function. This step-by-step approach ensures we account for all components of the composite function.

Simplification in Calculus

Simplifying complex expressions is essential for making derivatives more manageable to understand and use. By reducing expressions like \( \frac{2}{|2s+1| \sqrt{4s^2+4s}} \) to a cleaner form, we make calculations easier and avoid errors.Initially, simplifying requires performing algebraic manipulations, such as:- Squaring trinomials: \((2s + 1)^2 - 1 = 4s^2 + 4s\).- Factoring expressions under square roots: \( \sqrt{4s^2 + 4s} \rightarrow 2\sqrt{s(s+1)} \). The final simplification of \( \frac{2}{|2s + 1| \cdot 2 \sqrt{s(s + 1)}} \) yields \( \frac{1}{|2s + 1| \sqrt{s(s + 1)}} \), revealing a clear, concise derivative that is more efficient for computational purposes. Always strive for the simplest form to enhance both understanding and application.