Problem 14 - Calculus for Biology and Medicine

Question

In Problems 1-28, differentiate the functions with respect to the independent variable. $$ f(x)=\frac{\sqrt{x^{2}+1}}{2+\sqrt{x^{2}+1}} $$

Step-by-Step Solution

Answer

The derivative is \( f'(x) = \frac{2x}{(2+\sqrt{x^2 + 1})^2 \cdot \sqrt{x^2 + 1}} \).

Step 1: Identify the Quotient Rule

The function \( f(x) = \frac{\sqrt{x^2 + 1}}{2 + \sqrt{x^2 + 1}} \) is a quotient of two functions. Use the quotient rule, which states that if \( h(x) = \frac{u(x)}{v(x)} \), then the derivative \( h'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} \). Identify \( u(x) = \sqrt{x^2 + 1} \) and \( v(x) = 2 + \sqrt{x^2 + 1} \).

Step 2: Differentiate the Numerator

Find \( u'(x) \) where \( u(x) = (x^2 + 1)^{1/2} \). Use the chain rule: \( u'(x) = \frac{1}{2}(x^2 + 1)^{-1/2} \cdot 2x = \frac{x}{\sqrt{x^2 + 1}} \).

Step 3: Differentiate the Denominator

Find \( v'(x) \) where \( v(x) = 2 + \sqrt{x^2 + 1} \). The derivative of \( 2 + \sqrt{x^2 + 1} \) is the same as the derivative of \( \sqrt{x^2 + 1} \), which is \( \frac{x}{\sqrt{x^2 + 1}} \) using the chain rule as previously found.

Step 4: Apply the Quotient Rule

Substitute \( u(x) \), \( u'(x) \), \( v(x) \), and \( v'(x) \) into the quotient rule formula. This gives: \[ f'(x) = \frac{\left(\frac{x}{\sqrt{x^2 + 1}}\right)(2 + \sqrt{x^2 + 1}) - (\sqrt{x^2 + 1})\left(\frac{x}{\sqrt{x^2 + 1}}\right)}{(2+\sqrt{x^2 + 1})^2} \].

Step 5: Simplify the Numerator

Simplify the numerator: \( \frac{x(2 + \sqrt{x^2 + 1})}{\sqrt{x^2 + 1}} - x = \frac{2x}{\sqrt{x^2 + 1}} + x - x = \frac{2x}{\sqrt{x^2 + 1}} \). The numerator simplifies to \( 2x \).

Step 6: Write the Final Derivative

The derivative of the function is \[ f'(x) = \frac{2x}{(2+\sqrt{x^2 + 1})^2 \cdot \sqrt{x^2 + 1}} \].

Key Concepts

Quotient Rule

The quotient rule is essential when you need to differentiate a function that is the division of two functions. Think of it like this: if a function can be expressed as \( h(x) = \frac{u(x)}{v(x)} \), then differentiating it is not straightforward like the sum or product of functions. You need the quotient rule. It states: \[h'(x) = \frac{u'(x) \cdot v(x) - u(x) \cdot v'(x)}{[v(x)]^2} \]where:

• \( u(x) \) and \( v(x) \) are differentiable functions.
• \( u'(x) \) and \( v'(x) \) are their respective derivatives.

The trick lies in remembering that the name itself gives the idea. Differentiate the numerator and multiply it by the denominator, then subtract the result of the numerator multiplied by the derivative of the denominator. Finally, square the denominator to finish it off. This method effectively breaks down a complex problem into smaller, more manageable steps, helping illuminate the path to the solution.

Chain Rule

The chain rule is your go-to when you have a composite function. A composite function is simply one function within another. Like in this exercise, you see functions inside square roots, which require you to use this rule. The chain rule helps us peel back these layers smoothly. For a function \( f(x) = g(h(x)) \), the chain rule states that:\[f'(x) = g'(h(x)) \cdot h'(x)\]Breaking it down:

• Identify the "outer" function \( g \) and "inner" function \( h \).
• Find the derivative of the outer function while leaving the inner function unchanged.
• Multiply this by the derivative of the inner function.

Applying it to \( u(x) = \sqrt{x^2+1} \), the outer function \( g \) is the square root, and the inner function \( h \) is \( x^2 + 1 \). You first find the derivative of the square root function and then multiply it with the derivative of \( x^2 + 1 \), giving the result as \( \frac{x}{\sqrt{x^2+1}} \). This rule is a powerful tool because it enables differentiation across complex layers.

Derivative Simplification

Simplifying derivatives is crucial to avoid complex expressions that are hard to interpret or work with. After applying the quotient and chain rules, you might end up with an expression that seems daunting. The goal is to break it down into its simplest form. When simplifying:

• Look for common factors in numerators and denominators that can be cancelled.
• Combine like terms wherever possible.
• Simplify fractions by dividing both numerator and denominator by their greatest common divisor, if possible.

In the exercise, after applying the quotient rule, we worked through the expression \( f'(x) = \frac{x(2 + \sqrt{x^2+1})}{(2+\sqrt{x^2+1})^{2} \cdot \sqrt{x^2+1}} - x \). Notice how cancelling terms and simplifying fractions can help reduce this to \( \frac{2x}{(2+\sqrt{x^2+1})^2 \cdot \sqrt{x^2+1}} \). Simplification is not just for elegance; it reduces errors and makes future calculations easier. Consistent practice is key to mastering these techniques.