Problem 3 - Calculus for Biology and Medicine

Question

Find the first and the second derivatives of each function. $$ g(x)=\frac{1}{x+1} $$

Step-by-Step Solution

Answer

First derivative: \( g'(x) = -\frac{1}{(x+1)^2} \), second derivative: \( g''(x) = \frac{2}{(x+1)^3} \).

Step 1: Rewrite the Function

The given function is \[ g(x) = \frac{1}{x+1} \]To differentiate the function, it's often easier to rewrite it by using the property of exponents. Rewrite the function as: \[ g(x) = (x+1)^{-1} \]

Step 2: Find the First Derivative

To find the first derivative, apply the power rule, which states that the derivative of \( x^n \) is \( n \cdot x^{n-1} \). Differentiate \( g(x) = (x+1)^{-1} \):\[ g'(x) = -1 \cdot (x+1)^{-2} \cdot (1) = -\frac{1}{(x+1)^2} \]Here, we also apply the chain rule, multiplying by the derivative of the inside function \( x+1 \), which is 1.

Step 3: Find the Second Derivative

Now, find the second derivative by differentiating the first derivative \( g'(x) = -\frac{1}{(x+1)^2} \). Using the power rule and chain rule again:\[ g''(x) = \frac{d}{dx}\left(-\frac{1}{(x+1)^2}\right) \]Rewrite and differentiate:\[ g''(x) = 2 \cdot (x+1)^{-3} = \frac{2}{(x+1)^3} \] This is obtained by differentiating \( -1 \cdot (x+1)^{-2} \) and applying the chain rule.

Key Concepts

Power Rule for Derivatives

The power rule is a straightforward and essential concept in calculus, used to find the derivative of functions that are power of a variable. The rule states that if you have a function written in the form \( x^n \), where \( n \) is any real number, then the derivative \( f'(x) \) is given by \( n \cdot x^{n-1} \).

In this process, you're effectively multiplying the exponent by the base raised to one power less than the original expression.

• Example: For \( x^3 \), the derivative is \( 3x^2 \).
• Example: For \( x^{-1} \), the derivative is \( -1 \cdot x^{-2} \).

It is important to rewrite functions that are fractions in their exponent form, like \( (x+1)^{-1} \), before applying the power rule. This makes differentiation manageable and directly applicable, allowing you to easily find the derivative as seen in the original exercise.

Chain Rule for Complex Functions

In calculus, the chain rule is crucial for differentiating compositions of functions. It is particularly useful when functions are nested or involve expressions within another function. The chain rule helps you differentiate a composite function of the form \( f(g(x)) \).

The rule states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function times the derivative of the inner function itself: \[ \frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x). \]

Applying the chain rule ensures that all parts of a composite function are correctly differentiated. Let's consider the derivative \( g(x) = (x+1)^{-1} \). After applying the power rule, we arrive at \( -1 \cdot (x+1)^{-2} \).
Here, we treat \((x+1)\) as the inner function and its derivative \( g'(x) = 1 \) is the factor multiplying our result, ensuring that the differentiation is complete.
Being aware of when and how to apply the chain rule allows you to handle more intricate derivatives, like the ones in the exercise.

Understanding Exponents

Exponents are fundamental in mathematics and play a critical role when dealing with derivatives, often simplifying the process of finding them. An exponent denotes how many times a number, the base, is multiplied by itself.

• For example, \( x^2 \) means \( x \) multiplied by itself, \( x \cdot x \).
• If an exponent is negative, like \( x^{-1} \), it indicates a reciprocal, becoming \( \frac{1}{x} \).

In terms of differentiation, exponents significantly affect how we apply the power rule. They transform complex functions into manageable parts by shifting them to fit the power rule framework.

As seen in the original exercise, converting \( \frac{1}{x+1} \) into \( (x+1)^{-1} \) simplifies the derivation process, making it easier to apply the power and chain rules efficiently. Exponential properties are key in keeping your calculations error-free and methodical.