Question

Suppose you forgot the Quotient Rule for calculating \(\frac{d}{d x}\left(\frac{f(x)}{g(x)}\right) .\) Use the Chain Rule and Product Rule with the identity \(\frac{f(x)}{g(x)}=f(x)(g(x))^{-1}\) to derive the Quotient Rule.

Short answer

Question: Derive the Quotient Rule for finding the derivative of a function in the form \(\frac{f(x)}{g(x)}\), using the Chain Rule, Product Rule, and the given identity \(\frac{f(x)}{g(x)} = f(x)(g(x))^{-1}\). Answer: The Quotient Rule for finding the derivative of a function in the form \(\frac{f(x)}{g(x)}\) can be derived as follows: \(\frac{d}{dx}\left(\frac{f(x)}{g(x)}\right) = \frac{f'(x)g(x) - f(x)g'(x)}{g(x)^2}\)

Step-by-step solution

Rewrite function using the identity

According to the given identity, we can rewrite the function as \(h(x) = f(x)(g(x))^{-1}\).

Apply Product Rule

Next, we need to calculate the derivative of \(h(x)\). Since this function is the product of two functions, \(f(x)\) and \((g(x))^{-1}\), we will apply the Product Rule here. The Product Rule states that if \(u(x) = s(x)t(x)\), then the derivative is given by \(u'(x) = s'(x)t(x) + s(x)t'(x)\). Applying this rule, we get: \(h'(x) = f'(x)(g(x))^{-1} + f(x)\frac{d}{dx}(g(x))^{-1}\).

Apply Chain Rule

Now, we need to find the derivative of \((g(x))^{-1}\). This will require applying the Chain Rule, which states that if \(v(x) = r(t(x))\) then the derivative is given by \(v'(x) = r'(t(x))\cdot t'(x)\). Here, \(t(x) = g(x)\) and the outer function \(r(t) = t^{-1}\). To find the derivative, we first calculate the derivative of \(r(t)\). \(r'(t) = -t^{-2}\) Next, we apply the Chain Rule to find the derivative of \((g(x))^{-1}\): \(\frac{d}{dx}(g(x))^{-1} = -g(x)^{-2}\cdot g'(x)\)

Use the results from Step 3 in Step 2

We will substitute the derivative of \((g(x))^{-1}\) in the expression from Step 2: \(h'(x) = f'(x)(g(x))^{-1} + f(x)(-g(x)^{-2}\cdot g'(x))\)

Simplify

Finally, we will simplify the expression obtained in Step 4 as follows: \(h'(x) = \frac{f'(x)}{g(x)} - \frac{f(x)g'(x)}{g(x)^2}\) Now, to express this in the form of the Quotient Rule, we rewrite it so that both terms have the same denominator: \(h'(x) = \frac{f'(x)g(x) - f(x)g'(x)}{g(x)^2}\) Thus, we have derived the Quotient Rule: \(\frac{d}{dx}\left(\frac{f(x)}{g(x)}\right) = \frac{f'(x)g(x) - f(x)g'(x)}{g(x)^2}\)

Key concepts

Chain Rule

The Chain Rule is a fundamental concept in calculus that helps us take the derivative of a composite function. Imagine you have a function inside another function. The Chain Rule lets us "unravel" these layers to find the derivative.

Here's how it works: If you have a function like \( v(x) = r(t(x)) \), the Chain Rule states that the derivative \( v'(x) \) is the product of the derivative of the outer function \( r \) evaluated at \( t(x) \), and the derivative of the inner function \( t(x) \).

In mathematical terms, this is: \[ v'(x) = r'(t(x)) \cdot t'(x) \]

The Chain Rule is super helpful when dealing with expressions like \( (g(x))^{-1} \), where we treat \( g(x) \) as an inner function and \( t^{-1} \) as the outer one. When applying this rule, always be careful to differentiate both parts correctly.

Product Rule

The Product Rule is crucial when you're working with the product of two functions. This rule tells us how to differentiate the product of two functions accurately.

If you have two functions \( u(x) = s(x) \cdot t(x) \), the Product Rule says that the derivative \( u'(x) \) is:

• Derivative of the first function times the second function, plus
• First function times the derivative of the second function.

This can be written mathematically as: \[ u'(x) = s'(x) \cdot t(x) + s(x) \cdot t'(x) \]

In the Quotient Rule derivation, we use this formula to differentiate the product \( f(x) \cdot (g(x))^{-1} \). It's a solid method when breaking down complex functions into manageable pieces.

Derivative

A derivative represents the rate at which a function changes as its input changes. It's a core concept in calculus and is used to find slopes and rates of change in various contexts.

When you compute the derivative of a function \( f(x) \), you're essentially looking at how \( f \) behaves around each \( x \) value. Algebraically, this is described as the limit:

\[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \]

Derivatives can tell you if a function is increasing, decreasing, or if it's steady. In the context of the Quotient Rule, derivatives of \( f(x) \) and \( g(x) \) are pivotal to finding the derivative of their quotient.

Calculus

Calculus is a branch of mathematics that studies change and motion. It gives us a way to describe how things evolve over time or space.

Calculus is broadly divided into two areas:

• Differential Calculus, which focuses on derivatives and understanding rates of change.
• Integral Calculus, which is about finding areas, volumes, and accumulations.

Differential calculus helps us understand how quantities change, and is closely connected to the concept of a derivative.

Understanding calculus is key to solving real-world problems involving dynamic systems—like physics, engineering, and economics. It equips you with powerful tools like the Chain and Product Rules to analyze and manipulate functions effectively for problem-solving.