Problem 17 - Calculus for Biology and Medicine

Question

Use (4.12) to find the derivative of the inverse at the indicated point. Let \(f(x)=\ln (\sin x), 0

Step-by-Step Solution

Answer

The derivative is \(\frac{1}{\sqrt{3}}\).

Step 1: Understand the Problem

We need to find the derivative of the inverse function of \(f(x) = \ln(\sin x)\) at \(x = -\ln 2\). This involves understanding the relation between \(f\) and its inverse \(f^{-1}\) and applying the formula for the derivative of an inverse function.

Step 2: Write the Formula for Derivative of the Inverse

The derivative of the inverse function at a particular point, \(x = f(a)\), can be found using the formula: \( (f^{-1})'(x) = \frac{1}{f'(a)} \). Here \(a\) is such that \(f(a) = x\).

Step 3: Find \(a\) such that \(f(a) = -\ln 2\)

Solve the equation \(f(a) = \ln(\sin a) = -\ln 2\). This means \(\sin a = \frac{1}{2}\). Hence, \(a = \frac{\pi}{6}\) since it is in the interval \(0 < a < \frac{\pi}{2}\).

Step 4: Calculate the Derivative \(f'(x)\)

Compute \(f'(x)\) using the chain rule: \(f'(x) = \frac{d}{dx}[\ln(\sin x)] = \frac{1}{\sin x} \cdot \cos x = \cot x\).

Step 5: Evaluate \(f'(a)\)

Substitute \(a = \frac{\pi}{6}\) into \(f'(x)\): \(f'(\frac{\pi}{6}) = \cot \frac{\pi}{6} = \sqrt{3}\).

Step 6: Find Derivative of Inverse Function

Using the inverse derivative formula: \((f^{-1})'(-\ln 2) = \frac{1}{f'(\frac{\pi}{6})} = \frac{1}{\sqrt{3}}\).

Key Concepts

Inverse Function

Inverse functions are like mathematical mirrors. They reverse the process of a function and give us the original input that created a specific output. Imagine if a function was a machine that accepts numbers and transforms them into new numbers - the inverse function turns them back to the original.

To find the derivative of an inverse function, we use the formula:

• \((f^{-1})'(x) = \frac{1}{f'(a)}\),

where \(f(a) = x\). This formula helps in calculating how the inverse function changes with respect to changes in \(x\).

In our exercise, we have the function \(f(x) = \ln(\sin x)\) and are tasked to find the inverse's derivative. We determine that for \(x = -\ln 2\), \(a\) results in \( \sin a = \frac{1}{2} \) which is at \(a = \frac{\pi}{6}\). This allows us to use the formula and find the derivative of \(f^{-1}(x)\).

Chain Rule

The chain rule is a powerful tool in calculus for finding the derivative of composite functions. When you have a function inside another, like \(f(g(x))\), the chain rule states that the derivative is the product of the derivative of the outer function evaluated at the inner function and the derivative of the inner function.

It's written as:

• \(\frac{d}{dx} [f(g(x))] = f'(g(x)) \cdot g'(x)\).

In our problem, the chain rule lets us derive \(f(x) = \ln(\sin x)\) by breaking it down. Here, \(f(x) = \ln u\) where \(u = \sin x\). So first, we derive \(\frac{d}{du}[\ln u] = \frac{1}{u}\) and then \(\frac{d}{dx}[\sin x] = \cos x\). These combine to give us \(f'(x) = \frac{1}{\sin x} \cdot \cos x = \cot x\), allowing us to evaluate \(f'\) later at \(a = \frac{\pi}{6}\).

Trigonometric Functions

Trigonometric functions connect the angles and sides of triangles. They form the backbone for many calculus problems. In our exercise, the function involves \(\sin x\), one of these fundamental trigonometric functions.

In solving for the inverse function’s derivative, the sine function plays a crucial role. It's important to recall that within the first quadrant where \(0 < x < \frac{\pi}{2}\), \(\sin x\) is positive. Solving \(\sin a = \frac{1}{2}\) tells us \(a = \frac{\pi}{6}\), since it's within the stated domain.

Once we find \(a\), it bridges our understanding between the angle and ratio, allowing us to compute \(f'(a)\), essential for using the formula for inverse derivatives.

Logarithmic Differentiation

Logarithmic differentiation is a method used for finding the derivative of certain functions easily. This technique is particularly useful when dealing with complex products or quotients, and functions like \(\ln(\sin x)\) in our exercise.

To differentiate \(f(x) = \ln(\sin x)\), observe how the natural logarithm simplifies multiplicative terms. The derivation using logarithmic properties simplifies to applying the chain rule on this function smoothly. We use the fact that:

• \(\frac{d}{dx}[\ln(u)] = \frac{1}{u} \cdot \frac{du}{dx}\),

where \(u = \sin x\). This simplification initially leads to \(\frac{1}{\sin x}\), and multiplying by the derivative \(\cos x\) of \(\sin x\), results in \(f'(x) = \cot x\).

This expression, \(\cot x\), is then used to find the inverse derivative by substituting \(a = \frac{\pi}{6}\).