Question

In Exercises \(81-84,\) differentiate (a) with respect to \(x\) ( \(y\) is a function of \(x\) ) and (b) with respect to \(t(x\) and \(y\) are functions of \(t\) ). $$ 2 y^{2}-3 x^{4}=0 $$

Short answer

The first derivative of the equation with respect to \(x\) is \(y'=3\frac{x^{3}}{y}\) and with respect to \(t\) is \(\frac{dy}{dt} = 3\frac{x^{3}}{y}*\frac{dx}{dt}\)

Step-by-step solution

Differentiating with respect to \(x\)

Start by differentiating the provided equation with respect to \(x\). Fist, the term \(2y^{2}\) becomes \(4y*y'\) through application of chain rule. In the chain rule, each function is differentiated successively. So the derivative of an internal function, in this case \(y(x)\), is also considered. The term \(-3x^{4}\) becomes \(-12x^{3}\). Therefore, the derivative of the equation is:\(4y*y' - 12x^{3} = 0\). Rewrite the equation to solve for \(y'\) and it becomes:\(y'=\frac{12x^{3}}{4y}=3\frac{x^{3}}{y}\)

Differentiating with respect to \(t\)

Now, differentiate the equation considering \(t\) such that both \(x\) and \(y\) are a function of it. Again apply the chain rule. First, the term \(2y^{2}\) differentiates to \(4y*\frac{dy}{dt}\). Then, the term \(-3x^{4}\) differentiates to \(-12x^{3}*\frac{dx}{dt}\). Therefore, following differentiation, the equation becomes:\(4y*\frac{dy}{dt} - 12x^{3}*\frac{dx}{dt} = 0\). Solving for \(\frac{dy}{dt}\) yields,\(\frac{dy}{dt} = \frac{12x^{3}*\frac{dx}{dt}}{4y} = 3\frac{x^{3}}{y}*\frac{dx}{dt}\)

Key concepts

Chain Rule

The chain rule is a fundamental tool in calculus that allows us to compute the derivative of a composite function. In other words, it helps us understand how changes in one variable affect another variable when they are linked through a function.

For example, if a variable y depends on a variable u which itself depends on the variable x, then the chain rule states that the derivative of y with respect to x can be found by multiplying the derivative of y with respect to u by the derivative of u with respect to x, symbolically:
\[\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}\].

This concept can be particularly useful in implicit differentiation, where y is not explicitly defined as a function of x. In the given exercise, when differentiating the term \(2y^2\) with respect to x, y is treated as a function of x. Hence, the chain rule is applied to calculate the derivative, which results in \(4y \frac{dy}{dx}\). This method is instrumental when handling variables that are implicitly related, as is usually the case in real-world problems.

Differentiation with Respect to X

Differentiating with respect to x involves taking the derivative of a function that is defined in terms of x. When the function is explicit, differentiating is straightforward. However, when dealing with implicit functions, where y is defined in terms of x but not explicitly solvable for y, it requires implicit differentiation.

In practice, this means treating y as a function of x and applying the differentiation rules, including the chain rule where necessary. As seen in the provided step-by-step solution, when we take the derivative of \(2y^2\) with respect to x, we get \(4y \frac{dy}{dx}\) because y is assumed to be a function of x. This process extracts the relationship between x and y's rates of change without the need to solve y explicitly in terms of x. It's a powerful technique for solving equations that define one variable implicitly in terms of another.

Differentiation with Respect to T

When variables are expressed as functions of a third variable, such as time t, the process is called differentiation with respect to t. This scenario often arises in physics and engineering, where x and y may represent spatial coordinates that change over time.

The step-by-step solution illustrates how to differentiate an implicit function with respect to a parameter t. Each term is differentiated separately, taking into account how both x and y depend on t. In the exercise, the differentiation of \(2y^2\) with respect to t is \(4y \frac{dy}{dt}\), as y is a function of t, and similarly for x. The chain rule here is critical as it allows the computation of these derivatives while understanding that x and y are not independent but are both functions of the same parameter t. This is immensely useful in dynamics and other applications where tracing how multiple related quantities change over time is essential.