Problem 56 - Calculus Single Variable

Question

Suppose that \(f\) and \(g\) are twice differentiable. Calculate a formula for \((f / g)^{\prime \prime}\).

Step-by-Step Solution

Answer

The formula for \((f/g)''\) is \(\frac{g^2 \cdot f'' - 2g \cdot g' \cdot f' - g \cdot g'' \cdot f + 2f \cdot (g')^2}{g^3}\).

Step 1: Apply Quotient Rule

The formula for the first derivative of a quotient \( \left( \frac{f}{g} \right)' \) is given by the quotient rule: \[ \left( \frac{f}{g} \right)' = \frac{g \cdot f' - f \cdot g'}{g^2} \]. This will be crucial for computing the second derivative.

Step 2: Use Product Rule and Chain Rule

To find \( \left( \frac{f}{g} \right)'' \), employ the product and chain rules on the expression for the first derivative. The formula becomes:\[ \left( \frac{f}{g} \right)'' = \frac{d}{dx} \left( \frac{g \cdot f' - f \cdot g'}{g^2} \right) \].

Step 3: Differentiate the Numerator

Calculate the derivative of \(g \cdot f' - f \cdot g'\) using the product rule: \[ \frac{d}{dx} (g \cdot f') = g' \cdot f' + g \cdot f'' \] and \[ \frac{d}{dx} (f \cdot g') = f' \cdot g' + f \cdot g'' \]. Thus, the derivative of the numerator is \[ g' \cdot f' + g \cdot f'' - (f' \cdot g' + f \cdot g'') \].

Step 4: Simplify the Numerator Derivative

Simplify the expression obtained: \[ g' \cdot f' + g \cdot f'' - f' \cdot g' - f \cdot g'' = g \cdot f'' - f \cdot g'' \].

Step 5: Differentiate the Denominator

Differentiate \(g^2\) with respect to \(x\) to get \(2g \cdot g'\). Thus, the derivative of the denominator \(g^2\) is \(2g \cdot g'\).

Step 6: Apply the Quotient Rule Again

Using the derivative of the numerator from Step 4 and the derivative of the denominator from Step 5, reapply the quotient rule: \[ \left( \frac{g \cdot f' - f \cdot g'}{g^2} \right)' = \frac{(g \cdot f'' - f \cdot g'') \cdot g^2 - (g \cdot f' - f \cdot g') \cdot 2g \cdot g'}{(g^2)^2} \].

Step 7: Simplify the Result

Express the equation in a cleaner form: \[ \frac{g^2 (g \cdot f'' - f \cdot g'') - (g \cdot f' - f \cdot g') 2g \cdot g'}{g^4} \], which simplifies to: \[ \frac{g^2 \cdot f'' - 2g \cdot g' \cdot f' - g \cdot g'' \cdot f + 2f \cdot (g')^2}{g^3} \].

Key Concepts

Quotient Rule

When you have a function that is the ratio of two other functions, this is where the quotient rule comes into play. The quotient rule is used to differentiate fractions of the form \( \frac{f}{g} \). Given that both \(f\) and \(g\) are differentiable functions, the formula for the derivative of this division is:

• \( \left( \frac{f}{g} \right)' = \frac{g \cdot f' - f \cdot g'}{g^2} \).

When applying this rule, it is crucial to remember:

• The numerator involves the derivative of the top function (\(f'\)) times the bottom function minus the original top function (\(f\)) multiplied by the derivative of the bottom function (\(g'\)).
• The denominator is simply the square of the original bottom function \(g\).

This rule simplifies the differentiation of quotients by making your computations systematic and easy to handle.

Product Rule

The product rule is vital when differentiating a product of two differentiable functions, \(u(x)\) and \(v(x)\). For example, to differentiate an expression like \(g \cdot f'\), you apply the product rule. The rule is given by:

• \( (u \cdot v)' = u' \cdot v + u \cdot v' \).

To break this down:

• First, you take the derivative of the first function and multiply it by the second function.
• Next, you take the original first function and multiply it by the derivative of the second function.

Combining these results gives you the derivative of the entire product. The product rule helps manage the differentiation of complex products by dividing them into smaller, easier-to-handle parts.

Chain Rule

The chain rule is an essential tool in calculus whenever you're dealing with functions nested within one another, often called composite functions. Suppose we have a composition of two functions, \( h(x) = f(g(x)) \). The chain rule helps us find the derivative of this composite function. It states:

• \( (f(g(x)))' = f'(g(x)) \cdot g'(x) \).

To understand this better:

• First, differentiate the outer function \(f\), keeping the inner function \(g(x)\) as it is.
• Then, multiply this result by the derivative of the inner function \(g'(x)\).

This rule is central when differentiating expressions where one function is inside another, making itpossible to peel back the outer layers and find derivatives in a structured way.

Second Derivative

The second derivative, noted as \(f''(x)\), identifies the concavity of a function and how its slope changes over time. In simpler terms, it's the derivative of the derivative of a function. If you have already differentiated a function, calculating the second derivative involves differentiating the first derivative.When working with a quotient like \( \frac{f}{g} \), you have to apply the quotient rule again to the first derivative you found in order to get the second derivative.

• Applying this rule results in a formula that gives insight into the curvature of \( \frac{f}{g} \).
• It demonstrates how the rate of change of \( \frac{f}{g} \) evolves across given points.

The second derivative is indispensable for understanding more about a function's underlying behavior, especially in identifying inflection points and determining where a function can be considered concave upward or downward.