Problem 38 - Calculus for Biology and Medicine

Question

Assume that \(f(x)\) and \(g(x)\) are differentiable. Find \(\frac{d}{d x} f\left[\frac{1}{g(x)}\right]\)

Step-by-Step Solution

Answer

The derivative is \( f'\left( \frac{1}{g(x)} \right) \times \left(-\frac{g'(x)}{(g(x))^2}\right) \).

Step 1: Apply the Chain Rule

In order to differentiate the composite function \( f\left[ \frac{1}{g(x)} \right] \), we will use the chain rule. The chain rule states that the derivative of \( f(u(x)) \) is \( f'(u(x)) \cdot u'(x) \), where \( u(x) = \frac{1}{g(x)} \).

Step 2: Differentiate the Inner Function

Firstly, let's find the derivative of the inner function \( u(x) = \frac{1}{g(x)} \). Using the reciprocal rule, the derivative is: \[ u'(x) = -\frac{g'(x)}{(g(x))^2} \].

Step 3: Differentiate the Outer Function

Now, express the derivative of the composite function using the chain rule. Since \( f(u) \) is the outer function with \( u = \frac{1}{g(x)} \): \[ \frac{d}{dx} f\left( \frac{1}{g(x)} \right) = f'\left( \frac{1}{g(x)} \right) \times \frac{d}{dx} \left( \frac{1}{g(x)} \right) = f'\left( \frac{1}{g(x)} \right) \times \left(-\frac{g'(x)}{(g(x))^2}\right) \].

Key Concepts

Composite Function

Imagine having two functions inside each other, like a Russian nesting doll. This is what we call a "composite function". If you have two functions, say \( f(x) \) and \( g(x) \), a composite function combines them, for example, \( f(g(x)) \).

• If we consider \( g(x) \) as the inner function, it affects the input first.
• Then, this result becomes the input for \( f(x) \), the outer function.

In our exercise, we have \( f\left[ \frac{1}{g(x)} \right] \). This means we first transform \( x \) using \( g(x) \), turn it into its reciprocal, and then apply \( f \).
To differentiate a composite function, the chain rule comes in handy, helping us to handle these interlocking results effectively.

Derivative

A derivative measures how one quantity changes as another quantity changes. Think of it as calculating the slope or steepness of a curve at a certain point.
In mathematical terms, it offers insight into the rate of change of a function. If \( f(x) \) is your function, the derivative, denoted \( f'(x) \), shows how \( y \) (the output) varies as \( x \) (the input) moves a tiny bit.

• For a simple function like \( x^2 \), the derivative \( 2x \) shows the change in the output for each small move in \( x \)."
• For more complex functions, like our \( f\left[ \frac{1}{g(x)} \right] \), understanding derivatives requires tools like the chain rule, to break down the steps.

Using derivatives, we can find how fast it's changing, even when functions loop into each other like this.

Reciprocal Rule

The reciprocal rule is vital when dealing with fractions in calculus. When you have a function expressed as a reciprocal, like \( \frac{1}{g(x)} \), you use the reciprocal rule to find its derivative.
This rule is a specific version of the chain rule where the numerator is always 1. It simplifies the process and ensures the correct calculation of derivatives for expressions like \( \frac{1}{g(x)} \).

• For a reciprocal function \( u(x) = \frac{1}{g(x)} \), the derivative is \( u'(x) = -\frac {g'(x)}{(g(x))^2} \).
• This formula comes from understanding that the reciprocal changes direction (hence the negative sign) while also factoring in the derivative of \( g(x) \) itself.

In complex derivatives, recognizing when a function is a reciprocal helps in applying this rule correctly and simplifies solving steps further, like in our original exercise.