Question

\(F=f / g\) be the quotient of two functions that are differentiable at \(x\) a. Use the definition of \(F^{\prime}\) to show that $$ \frac{d}{d x}\left(\frac{f(x)}{g(x)}\right)=\lim _{h \rightarrow 0} \frac{f(x+h) g(x)-f(x) g(x+h)}{h g(x+h) g(x)} $$ b. Now add \(-f(x) g(x)+f(x) g(x)\) (which equals 0 ) to the numerator in the preceding limit to obtain $$\lim _{h \rightarrow 0} \frac{f(x+h) g(x)-f(x) g(x)+f(x) g(x)-f(x) g(x+h)}{h g(x+h) g(x)}$$ Use this limit to obtain the Quotient Rule. c. Explain why \(F^{\prime}=(f / g)^{\prime}\) exists, whenever \(g(x) \neq 0\)

Short answer

In summary, the Quotient Rule for derivatives states that the derivative of a quotient function \(F(x) = \frac{f(x)}{g(x)}\) is given by: $$ F^{\prime}(x) = \frac{f^{\prime}(x)g(x)-g^{\prime}(x)f(x)}{g(x)^2} $$ The derivative of this quotient exists as long as the denominator function \(g(x) \neq 0\). This is because if \(g(x) = 0\), the denominator in the Quotient Rule would be equal to zero, causing the expression for the derivative of F to be undefined. Since both f and g are differentiable at x, and g(x) is nonzero, it guarantees that the derivative of the quotient of the functions exists.

Step-by-step solution

Write out the definition of the derivative and manipulate the expression

Recall the definition of the derivative of a function F at a point x is given by: $$ F^{\prime}(x)=\lim_{h \rightarrow 0}\frac{F(x+h)-F(x)}{h} $$ Substitute \(F(x) = \frac{f(x)}{g(x)}\), then the derivative of F would be: $$ F^{\prime}(x)=\lim_{h \rightarrow 0}\frac{\frac{f(x+h)}{g(x+h)}-\frac{f(x)}{g(x)}}{h} $$ To simplify the expression, find a common denominator for the fractions in the numerator: $$ F^{\prime}(x) = \lim_{h \rightarrow 0}\frac{f(x+h)g(x)-f(x)g(x+h)}{h g(x+h) g(x)} $$

Add 0 in the form of \(-f(x)g(x) + f(x)g(x)\) to the numerator

Now, we will add 0 to the numerator in the form of \(-f(x)g(x) + f(x)g(x)\): $$ F^{\prime}(x) = \lim_{h \rightarrow 0}\frac{f(x+h)g(x)-f(x)g(x)+f(x)g(x)-f(x)g(x+h)}{h g(x+h) g(x)} $$

Simplify the expression and derive the Quotient Rule

To obtain the Quotient Rule, notice that the numerator can be rearranged as the difference of two products: $$ F^{\prime}(x) = \lim_{h \rightarrow 0}\frac{[f(x+h)g(x)-f(x)g(x)]-[f(x)g(x+h)-f(x)g(x)]}{h g(x+h) g(x)} $$ Now, factor out \(f(x)\) and \(f(x+h)\) from the two terms in the numerator: $$ F^{\prime}(x) = \lim_{h \rightarrow 0}\frac{f(x+h)[g(x)-g(x+h)]-f(x)[g(x)-g(x+h)]}{h g(x+h) g(x)} $$ Factoring further, we get: $$ F^{\prime}(x) = \lim_{h \rightarrow 0}\frac{[f(x+h)-f(x)][g(x)-g(x+h)]}{h g(x+h) g(x)} $$ From the definition of a derivative, we can rewrite the limit in the numerator: $$ F^{\prime}(x) = \frac{f^{\prime}(x)[g(x)-g(x+h)]-g^{\prime}(x)[f(x+h)-f(x)]}{g(x)g(x+h)} $$ Taking the limit as \(h \rightarrow 0\) gives: $$ F^{\prime}(x) = \frac{f^{\prime}(x)g(x)-g^{\prime}(x)f(x)}{g(x)^2} $$ This expression is the Quotient Rule for derivatives.

Explain the conditions for the existence of the derivative

The derivative \(F^{\prime}(x)=(f/g)^{\prime}(x)\) exists whenever \(g(x) \neq 0\), because if \(g(x) = 0\), the denominator in the Quotient Rule would be equal to zero, causing the expression for the derivative of F to be undefined. Since f and g are both differentiable at x, and g(x) is nonzero, it guarantees that the derivative of the quotient of the functions exists.

Key concepts

Derivative

The concept of a derivative is central in calculus as it represents how a function changes at any point. Specifically, it describes the rate of change or slope of the function at a particular point. When we say we are "differentiating" a function, we are computing its derivative. This process unveils how the function's values evolve, as the input variables undergo small changes.

In mathematical terms, the derivative of a function \( F \), evaluated at a point \( x \), is expressed through the limit:

• \( F^{\prime}(x) = \lim_{h \rightarrow 0} \frac{F(x+h) - F(x)}{h} \)

This formula essentially computes the slope of the tangent line to the curve representing the function at the point \( x \).
For functions defined as a quotient of two differentiable functions, such as \( F = \frac{f}{g} \), the derivative becomes more complex. By applying the Quotient Rule, which is derived from the limit definition, it allows us to find the derivative:

• \( F^{\prime}(x) = \frac{f^{\prime}(x)g(x) - f(x)g^{-prime}(x)}{g(x)^2} \)

This rule is essential for differentiating expressions that are fractions of functions.

Differentiability

Differentiability refers to the condition whereby a function has a derivative at all points in its domain. In simpler terms, if you can draw a tangent line to the curve of the function at any given point, then it's differentiable at that point. This means the function behaves smoothly and doesn't have any sharp corners or discontinuities at that point.

For a function to be differentiable at a specific point, it must satisfy three critical conditions:

• The function must be continuous at that point.
• The limit of the derivative as \( h \rightarrow 0 \) must exist.
• The result of the derivative must be an actual number, not infinite.

Considering a quotient of two functions, \( F = \frac{f}{g} \), both \( f \) and \( g \) must be differentiable functions at \( x \). Additionally, the denominator \( g(x) \) should not be zero to avoid undefined expressions. These conditions ensure the smoothness of the function, ensuring the derivative \( F^{\prime}(x) \) exists.
Ultimately, differentiability ensures that functions can be analyzed and worked with effectively using calculus techniques, providing insights into rates of change and the behavior of functions.

Limit

The concept of a limit is fundamental in calculus and pivotal when discussing derivatives and differentiability. It defines the value that a function approaches as the input approaches a certain point. Limits help us understand how functions behave both at points and as inputs approach particular values.
In the context of the derivative and the Quotient Rule (used here for differentiating \( F = \frac{f}{g} \)), the limit is crucial. The derivative relies on:

• \( F^{\prime}(x) = \lim_{h \rightarrow 0} \frac{\frac{f(x+h)}{g(x+h)} - \frac{f(x)}{g(x)}}{h} \)

This expression is a limit that captures the idea of finding the slope of the curve at \( x \) by considering very small changes \( h \) in distance.
Limits also enable the manipulation needed for simplifying expressions, such as when adding the term \(-f(x)g(x) + f(x)g(x)\) during the derivation of the Quotient Rule. This manipulation, underpinned by the properties of limits, assures correctness and viability in deriving the final form of the Quotient Rule.
Ultimately, limits allow us to understand the behavior of functions around specific points, helping us define and calculate derivatives comprehensively.