Question

Derivatives Find and simplify the derivative of the following functions. $$h(w)=\frac{w^{2}-1}{w^{2}+1}$$

Short answer

Question: Find the simplified derivative of the function $$h(w)=\frac{w^{2}-1}{w^{2}+1}$$. Answer: The simplified derivative of the given function is $$h'(w) = \frac{4w}{(w^{2}+1)^2}$$.

Step-by-step solution

Identify the function as a quotient of two functions

The given function $$h(w)=\frac{w^{2}-1}{w^{2}+1}$$ is in the form \(\frac{u(w)}{v(w)}\), where $$u(w) = w^{2}-1$$ and $$v(w) = w^{2}+1$$.

Apply the quotient rule for differentiation

The quotient rule states that if $$h(w)=\frac{u(w)}{v(w)}$$, then the derivative of $$h(w)$$ with respect to $$w$$ is given by: $$h'(w)=\frac{u'(w)v(w)-u(w)v'(w)}{[v(w)]^2}$$. Now, let's find the derivatives of $$u(w)$$ and $$v(w)$$ with respect to $$w$$: $$u'(w) = \frac{d}{dw}(w^{2}-1) = 2w$$ and $$v'(w) = \frac{d}{dw}(w^{2}+1) = 2w$$. Next, apply the quotient rule: $$h'(w) = \frac{(2w)(w^{2}+1)-(w^{2}-1)(2w)}{[w^{2}+1]^2}$$.

Simplify the expression obtained after differentiating

We now need to simplify the expression for the derivative: $$h'(w) = \frac{2w^3+2w-2w^3+2w}{[w^{2}+1]^2} = \frac{4w}{(w^{2}+1)^2}$$. So, the simplified derivative of the given function is: $$h'(w) = \frac{4w}{(w^{2}+1)^2}$$.

Key concepts

Quotient Rule for Differentiation

The quotient rule for differentiation is an essential calculus technique used when taking the derivative of a function that is the ratio of two other functions. It's specifically tailored for the situation where one function is divided by another. Let's take a closer look at this rule to understand why it's so vital in calculus.

The quotient rule formula is expressed as follows: If you have a function in the form of \(h(w) = \frac{u(w)}{v(w)}\), then the derivative, denoted by \(h'(w)\), can be calculated using the quotient rule which tells us that \(h'(w) = \frac{u'(w)v(w) - u(w)v'(w)}{[v(w)]^2}\). It's crucial to note that in this formula, \(u'(w)\) and \(v'(w)\) signify the derivatives of \(u\) and \(v\), respectively, with respect to \(w\).

For the given exercise, the function \(h(w)\) is the quotient of \(w^2 - 1\) and \(w^2 + 1\). When applying the quotient rule, differentiation becomes a systematic process where we differentiate the numerator and the denominator separately, then put these pieces together as specified by the quotient rule formula. By mastering this rule, students can tackle complex differentiation problems with confidence. It's all about recognizing the form of the function and methodically applying the rule.

Simplifying Derivatives

Simplifying derivatives is a key step that follows the mechanical process of applying differentiation rules, such as the quotient rule. This process ensures that the derivative expression is neat, manageable, and as simple as possible, which is critical for further mathematical manipulation or interpretation of the function's behavior.

In the context of our exercise, after applying the quotient rule for differentiation to the function \(h(w)\), we obtain an expression for the derivative that may initially look complex. However, we can simplify the result by combining like terms and factoring out common factors.

For example, after applying the quotient rule to \(h(w)\), we have an expression that includes terms with \(w^3\) which cancel each other out, and \(2w\) terms that can be combined. Finally, after eliminating the unnecessary parts, the resultant expression is much cleaner: \(h'(w) = \frac{4w}{(w^2+1)^2}\). This simplified form is not only easier to read and understand but also more useful if we need to evaluate the derivative at particular values of \(w\) or analyze the function's behavior. Therefore, the simplification step is just as crucial as the differentiation itself.

Calculus

Calculus, the branch of mathematics that studies continuous change, is an indispensable tool in various scientific fields such as physics, engineering, economics, statistics, and many more. It is composed of two fundamental concepts: differentiation and integration, which are the reverse processes of each other.

In the context of solving problems, calculus not only provides the machinery to find derivatives and integrals but also the framework to interpret these operations. Differentiation, as portrayed through the quotient rule, allows us to find the rate at which a quantity changes. Meanwhile, integration helps in finding the quantity accumulated over time or space, given its rate of change.

The exercise we discussed focuses on differentiation, specifically using the quotient rule to find the derivative. Through this example, students can see how calculus enables us to take an initially complicated-looking function and, by differentiating it, gleam information about its rate of change. This type of analysis is vital in everything from designing a roller coaster's path to optimizing investments, thereby highlighting the ubiquitous and powerful nature of calculus in problem-solving.