Question

Using the Alternative Form of the Derivative In Exercises \(65-74,\) use the alternative form of the derivative to find the derivative at \(x=c\) (if it exists). $$ g(x)=(x+3)^{1 / 3}, \quad c=-3 $$

Short answer

The derivative of \(g(x) = (x+3)^{1 / 3}\) at \(x = -3\) does not exist.

Step-by-step solution

Identify the function

The function to differentiate is \(g(x) = (x+3)^{1/3}\) and the point of interest is at \(x = -3\).

Apply the Alternative Form of Derivative Formula

The alternative form of the derivative is \[ g'(c) = \lim_{{x \to c}} \frac{{g(x) - g(c)}}{{x - c}} \] So, apply the formula by replacing \(g(x)\) with \((x+3)^{1/3}\) and \(c\) with \(-3\). The result is \[ g'(-3) = \lim_{{x \to -3}} \frac{{(x+3)^{1/3} - (-3+3)^{1/3}}}{{x - (-3)}} \] This simplifies to \[ g'(-3) = \lim_{{x \to -3}} \frac{{(x+3)^{1/3}}}{x + 3} \]

Compute the limit

This limit is a 0/0 indeterminate form that can be resolved using l'Hopital's Rule, which states that \[ \lim_{{x \to c}} \frac{{f(x)}}{{g(x)}} = \lim_{{x \to c}} \frac{{f'(x)}}{{g'(x)}} \] if the limit exists. By differentiating both the numerator and the denominator of the fraction and finding the limit again, we get \[ g'(-3) = \lim_{{x \to -3}} \frac{{1/3*(x+3)^{-2/3}}}{1} \] which simplifies to \[ g'(-3) = \frac{1}{3 * (-3+3)^{2/3}} \]

Evaluate the derivative at \( x = -3 \)

Substitute \(x = -3\) into the simplified expression to evaluate the derivative at the given point. This yields \[ g'(-3) = \frac{1}{3 * 0^{2/3}} = undefined. \] Hence, the derivative of the function \(g(x) = (x+3)^{1 / 3}\) at \(x = -3\) does not exist.

Key concepts

Derivative at a Point

The derivative of a function at a specific point provides us with the slope of the tangent line to the graph of the function at that point. Understanding the derivative at a point is crucial for analyzing instantaneous rates of change, which is the essence of differential calculus.

For the function in question, we are interested in finding the derivative of the function \(g(x) = (x+3)^{1/3}\) at \(x=c\), where \(c=-3\). We examine the behavior of \(g(x)\) as \(x\) approaches the value of \(c\) using the alternative form of the derivative. This approach gives us precise information about the function's behavior exactly at the point \(x=c\) if the derivative exists there.

L'Hopital's Rule

L'Hopital's Rule is an essential tool for evaluating limits that present an indeterminate form, such as \(0/0\) or \(\infty/\infty\). When the straightforward substitution in a limit expression yields an indeterminate form, L'Hopital's Rule allows us to compute the limit by differentiating the numerator and the denominator separately and then taking the limit of their quotient.

In our derivative problem, applying L'Hopital's Rule is crucial because we have an indeterminate form when substituted into the limit expression. By taking the derivatives of the numerator and denominator and evaluating the limit, we attempt to find a conclusive result. However, the rule only applies if the limit of the derivatives leads to a determinate form or infinite limit.

Indeterminate Forms

Indeterminate forms occur when the limit of an expression is not immediately evident and could potentially represent several different values. The most common indeterminate forms are \(0/0\), \(\infty/\infty\), \(0 \times \infty\), \(\infty - \infty\), \(0^0\), \(\infty^0\), and \(1^\infty\).

Our exercise features the indeterminate form \(0/0\), as the expression presented after substituting \(x=c\) into the alternative form of the derivative yields zero in both the numerator and the denominator. This form requires a deeper analysis, like the application of L'Hopital's Rule, to determine the limit's value.

Limits in Calculus

Limits are foundational to calculus and are used to describe the behavior of functions as they approach a specific value. They are essential in defining derivatives, integrals, and continuity. A limit can either exist and equal a finite number, be infinite, or not exist.

In the context of our exercise, we analyze the limit of \(g(x)\) as \(x\) approaches \(-3\). The goal is to discover the behavior of \(g(x)\) at this point and, by extension, the derivative of \(g(x)\) at \(x = -3\). Unfortunately, even after employing L'Hopital's Rule, we find that the limit leads to division by zero, indicating that the derivative at this point does not exist. This highlights the importance of carefully evaluating limits and recognizing cases where a function may not have a derivative at certain points.