Problem 2 - Calculus for Biology and Medicine

Question

Find the inverse of each function and differentiate each inverse in two ways: (i) Differentiate the inverse function directly, and (ii) use (4.14) to find the derivative of the inverse. $$ f(x)=\sqrt{x+1}, x \geq-1 $$

Step-by-Step Solution

Answer

Inverse: \( f^{-1}(x) = x^2 - 1 \); Derivatives: Direct \( 2x \), Using (4.14) \( 2x \).

Step 1: Find the Inverse Function

To find the inverse of the function \( f(x) = \sqrt{x+1} \), swap \(x\) and \(y\), which represents \( f(x) \). Rearranging, we get \( x = \sqrt{y+1} \). Square both sides to eliminate the square root: \( x^2 = y+1 \). Solving for \( y \), we find the inverse function: \( y = x^2 - 1 \). Thus, the inverse function is \( f^{-1}(x) = x^2 - 1 \).

Step 2: Differentiate the Inverse Directly

Differentiate \( f^{-1}(x) = x^2 - 1 \) with respect to \( x \). Using the power rule, the derivative of \( x^2 \) is \( 2x \) and the constant \(-1\) differentiates to 0, so \( (f^{-1})'(x) = 2x \).

Step 3: Differentiate the Original Function

Differentiate the original function \( f(x) = \sqrt{x+1} \). Rewrite as \( f(x) = (x+1)^{1/2} \) and apply the chain rule: \( f'(x) = \frac{1}{2}(x+1)^{-1/2} \cdot 1 \), simplifying to \( \frac{1}{2\sqrt{x+1}} \).

Step 4: Use Formula (4.14) to Differentiate the Inverse

Formula (4.14) states \((f^{-1})'(y) = \frac{1}{f'(x)}\) where \(y = f(x)\). Use \( y = \sqrt{x+1} \) to solve for \(x\) in terms of \(y\): \(x = y^2 - 1\). Then evaluate \( f'(x) \) at \( x = y^2 - 1 \), resulting in \( f'(y^2 - 1) = \frac{1}{2\sqrt{y^2}} = \frac{1}{2y} \). The inverse derivative is \((f^{-1})'(y) = \frac{1}{2y} \). Substitute \( x \) for \( y \) since \( y = x \) for the inverse, resulting in \((f^{-1})'(x) = 2x \).

Key Concepts

Derivatives

A derivative is a fundamental concept in calculus that represents the rate of change of a function as its input changes. This is like measuring how fast a car is traveling at any given time. In simple terms, the derivative tells you the slope of the function at any point on its graph.
For example, if you have a function like \( f(x) = x^2 \), its derivative \( f'(x) = 2x \) gives you the slope at any specific point on its curve. The derivative is particularly useful for understanding how functions behave and helps with optimization problems, where we need to find maxima or minima.

• To differentiate a function, identify the function's formula first.
• Apply differentiation rules such as the power rule or chain rule as necessary.
• Once you have the derivative, you can solve many real-world problems.

Understanding derivatives lays the groundwork for exploring more complex functions, like those with higher degrees or composite forms.

Chain Rule

The chain rule is an essential technique used in calculus when dealing with the composite functions. A composite function is like combining two functions into one, for instance, \( g(x) = (h(x))^2 \). The chain rule allows us to differentiate these functions with ease.
The chain rule states that if you have a function \( f(g(x)) \), the derivative \( f'(g(x)) \) is found by multiplying the derivative of the outer function \( f \) evaluated at \( g(x) \) by the derivative of the inner function \( g'(x) \). In notation: \( f'(g(x)) = f'(g) \cdot g'(x) \).

• First, identify the inner and outer functions.
• Differentiate the outer function and leave the inner function as it is.
• Then, differentiate the inner function.
• Multiply the results.

When applied correctly, the chain rule simplifies the differentiation process for complex nested functions, making it an invaluable tool.

Power Rule

The power rule is one of the simplest and most commonly used rules in calculus for differentiating functions. If you have a power function of the form \( x^n \), the power rule states that the derivative is \( nx^{n-1} \).
For instance, from \( f(x) = x^3 \), the derivative using the power rule would be \( 3x^2 \). This is because you multiply by the power and then reduce the power by one.

• The power rule is easy to apply, even for terms with large exponents.
• It is a quick way to differentiate functions expressed as simple polynomial terms.
• It can be used in combination with other rules, like the chain rule, for more complex functions.

Mastering the power rule is crucial for efficiently finding derivatives of simple and complex expressions, forming a baseline for tackling integration in advanced calculus.