Problem 4 - Calculus Single Variable

Question

\( y\) is a function of \(x .\) Calculate the derivative of the given expression with respect to \(x\). (Your answer should contain the term \(d y / d x .)\) \(\left(y^{3}-1\right) / y\)

Step-by-Step Solution

Answer

The derivative of \(\frac{y^3 - 1}{y}\) is \(2y \frac{dy}{dx} + \frac{1}{y^2} \frac{dy}{dx}\).

Step 1: Rewrite the Expression

First, rewrite the given expression \(\frac{y^3 - 1}{y}\) as \(y^2 - \frac{1}{y}\). This simplification makes it easier to differentiate.

Step 2: Differentiate Each Term

Differentiate the expression \(y^2 - \frac{1}{y}\) with respect to \(x\). Recall that using the chain rule, the derivative of \(y^n\) is \(n y^{n-1} \frac{dy}{dx}\).- Derivative of \(y^2\) is \(2y \frac{dy}{dx}\).- Derivative of \(-\frac{1}{y}\) is \(\frac{1}{y^2} \frac{dy}{dx}\) using the chain rule.

Step 3: Combine the Derivatives

Combine the derivatives from the previous step:The derivative of the expression is \(2y \frac{dy}{dx} + \frac{1}{y^2} \frac{dy}{dx}\).

Key Concepts

Chain Rule

The chain rule is an essential tool in calculus, particularly when dealing with composite functions. It allows us to differentiate functions that are nested inside each other, such as in the form of \(f(g(x))\). The main idea is to take the derivative of the outer function and multiply it by the derivative of the inner function.

In the context of differentiating \(y^2 - \frac{1}{y}\), the chain rule is applied to functions of \(y\) when you express them in terms of \(x\). This is because when you differentiate \(y^n\) with respect to \(x\), you must account for how \(y\) changes with \(x\), hence the inclusion of \(\frac{dy}{dx}\).

• To differentiate \(y^2\), recognize it as a power function. Apply the power rule: bring down the exponent and reduce it by one. This gives \(2y\), but since \(y\) depends on \(x\), you multiply by \(\frac{dy}{dx}\) using the chain rule. Thus, the derivative is \(2y \frac{dy}{dx}\).
• The term \(-\frac{1}{y}\) involves \(y^{-1}\). Differentiating involves the power rule too: the power of \(y\) decreases by one, rendering \(-1 y^{-2}\). Again, multiply by \(\frac{dy}{dx}\). Therefore, its derivative is \(\frac{1}{y^2} \frac{dy}{dx}\).

Differentiation

Differentiation is the process of finding a derivative, which is essentially the rate at which a function changes at any given point. It's one of the core operations in calculus and is used extensively in many areas, such as physics, engineering, and economics.

When you differentiate a function, you are trying to find how the y-value (output) changes as the x-value (input) changes. This change is reflected in the slope of the tangent line to the function at any given point.

• For polynomial terms like \(y^2\), use the power rule which states that the derivative of \(x^n\) is \(n \, x^{n-1}\).
• For rational functions like \(-\frac{1}{y}\), first express it as \(y^{-1}\), so you can apply the power rule easily.
• Both cases observed in the exercise require consideration of how changes in \(y\) influence the derivative with respect to \(x\), which is captured by incorporating \(\frac{dy}{dx}\).

Implicit Differentiation

Implicit differentiation is a method used when a function is not easily solved for one variable in terms of the other. This becomes necessary when the function is given implicitly rather than explicitly.

Typically, implicit differentiation involves treating one variable as a function of another. In the exercise, \(y\) is considered a function of \(x\) even though it's not explicitly solved as \(y=f(x)\). The derivative \(\frac{dy}{dx}\) is included in every differentiation step involving \(y\).

• Assume that \(y\) is a function of \(x\). This assumption is key to applying implicit differentiation.
• While differentiating each term with respect to \(x\), attach \(\frac{dy}{dx}\) whenever you differentiate a \(y\) term, acknowledging \(dy/dx\) as the derivative of \(y\) with respect to \(x\).
• This method allows you to differentiate complex expressions involving variables that cannot be isolated easily.