Question

By means of the definition of \(\$ 1\) differentiate each of the following functions: \(y=\frac{1}{x^{3}}\)

Short answer

Question: Find the derivative of the function \(y=\frac{1}{x^3}\) using the definition of the derivative. Answer: The derivative of the function \(y=\frac{1}{x^3}\) is \(y' = -3x^{-4}\).

Step-by-step solution

Write down the definition of the derivative

The definition of the derivative of a function \(f(x)\) with respect to \(x\) is given by: $$f'(x) = \lim_{h\to 0} \frac{f(x+h) - f(x)}{h}$$

Substitute the given function into the definition

We are given the function \(y=\frac{1}{x^3}\). Substitute this function into the definition of the derivative: $$y'(x) = \lim_{h\to 0} \frac{\frac{1}{(x+h)^3} - \frac{1}{x^3}}{h}$$

Simplify the expression

To simplify the expression in the numerator, find a common denominator for the two fractions and then combine them: $$y'(x) = \lim_{h\to 0} \frac{x^3 - (x+h)^3}{h(x^3)(x+h)^3}$$

Expand the binomial

We need to expand \((x+h)^3\): $$y'(x) = \lim_{h\to 0} \frac{x^3 - (x^3 + 3x^2h + 3xh^2 + h^3)}{h(x^3)(x+h)^3}$$

Simplify the expression further

Now, we can cancel out the \(x^3\) term in the numerator: $$y'(x) = \lim_{h\to 0} \frac{-3x^2h - 3xh^2 - h^3}{h(x^3)(x+h)^3}$$

Factor and cancel

Factor out the \(h\) in the numerator and cancel out the term with the \(h\) in the denominator: $$y'(x) = \lim_{h\to 0} \frac{h(-3x^2 - 3xh - h^2)}{h(x^3)(x+h)^3}$$ $$y'(x) = \lim_{h\to 0} \frac{-3x^2 - 3xh - h^2}{x^3(x+h)^3}$$

Take the limit as h approaches 0

Now, take the limit as \(h\) approaches \(0\): $$y'(x) = \frac{-3x^2 -3x(0) - (0)^2}{x^3(x)^3}$$

Simplify the final expression

Finally, we simplify the expression to get the derivative of the function: $$y'(x) = \frac{-3x^2}{x^6}$$ $$y'(x) = -3x^{-4}$$ Thus, the derivative of the given function \(y=\frac{1}{x^3}\) is \(y' = -3x^{-4}\).

Key concepts

Definition of Derivative

Understanding the definition of the derivative is crucial when dealing with calculus problems. A derivative represents the rate at which one quantity changes with respect to another. To put it simply, if you have a function that describes how a position changes over time, the derivative of that function tells you the velocity—the rate of change of position over time.

In mathematical terms, the derivative of a function at a specific point is the limit of the average rate of change of the function over an interval as the interval becomes infinitesimally small. The formal definition is given by: \[f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}\].
This definition is derived using the concept of the limit of a function, which is another fundamental element of calculus. By calculating the derivative, we are essentially finding the instantaneous rate of change at a specific point on the function's graph.

Limit of a Function

The concept of the limit underpins much of calculus and helps explain what happens to a function as we get incredibly close to a specific point. A limit asks the question: 'As the input of the function approaches some value, what value does the output of the function approach?'

The notation \[@\alpha=(x)\] indicates the value that \(f(x)\) gets closer to as \(x\) approaches \(a\). Limits are not always straightforward and can sometimes be undefined or approach infinity. They are especially useful when you cannot evaluate a function directly at a particular point, typically because it would lead to an indeterminate form like 0/0. The process of finding the limit as \(h\) approaches zero in the derivative computation is what allows us to calculate the instantaneous rate of change without needing infinitesimally small intervals in real measurements.

Simplifying Expressions

Simplifying expressions is a frequently used skill in calculus, and it's often the first step in solving more complex mathematical problems. It primarily involves reducing fractions to their lowest terms, combining like terms, and factoring. This process makes it easier to carry out further operations such as differentiation or integration.

When simplifying the expression for a derivative, for example, finding a common denominator and factoring allows you to cancel terms and often reduce the complexity of the arithmetic involved. In the exercise on differentiating the inverse cubic function, we factored out terms and canceled the \(h\) in the numerator with the \(h\) in the denominator, which allowed us to continue the calculation and eventually reach a simpler form for the derivative of the function.

Binomial Expansion

Binomial expansion is involved in many calculus problems, especially when dealing with polynomial functions. It allows us to expand expressions of the form \((a + b)^n\), where \(n\) is a positive integer. The expanded form is a sum of terms of the form \(C(n, k)a^{n-k}b^k\), where \(C(n, k)\) denotes the binomial coefficient, calculated as \(\frac{n!}{k!(n-k)!}\).

In the context of differentiating inverse cubic functions, binomial expansion lets us rewrite the term \((x + h)^3\) as \(x^3 + 3x^2h + 3xh^2 + h^3\). This expanded form is essential because it allows us to cancel similar terms when simplifying the expression for the derivative.
Understanding these concepts and their relationships can significantly enhance the problem-solving skills of any student trying to master calculus. By breaking down the problem and working through each concept methodically, as demonstrated in the provided exercise, students can gain confidence and improve their mathematical understanding.