91Ó°ÊÓ

When we say that a function is differentiable, it means that the function has a derivative. The derivative of a function measures how the function's output value changes as its input value changes. In other words, it gives us the rate of change of the function with respect to its variable.

Differentiable functions have several important properties:

• They are continuous, meaning there are no abrupt jumps or holes in their graphs.
• They are smooth, meaning they do not have sharp corners or cusps.

In the given exercise, both variables, \(x\) and \(y\), are differentiable functions of \(t\). This means we can find their derivatives with respect to \(t\), which is essential for applying the rules of differentiation.

The Chain Rule is a fundamental technique in calculus for finding the derivative of a composite function. It helps us differentiate a function based on the derivatives of its inner functions. Formally, if we have a function \(y = f(u)\) and \(u = g(x)\), the Chain Rule states that:

\[\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}\]

In the context of our exercise, when we differentiate \(x^4 + y^4 = 1\) with respect to \(t\), we need to apply the Chain Rule to both \(x^4\) and \(y^4\) since \(x\) and \(y\) themselves are functions of \(t\). For example, when differentiating \(x^4\), we use:

\[\frac{d}{dt} x^4 = 4x^3 \cdot \frac{dx}{dt}\].

This allows us to find how \(x\) and \(y\) change with respect to \(t\) even when their relationship to \(t\) is implicit.

The Power Rule is another essential rule in differentiation. It states that if we have a function \(f(x) = x^n\), where \(n\) is any real number, its derivative is:

\[\frac{d}{dx} x^n = n x^{n-1}\]

In our exercise, we make use of the Power Rule to differentiate terms like \(x^4\) and \(y^4\). To differentiate \(x^4\) with respect to \(t\), we notice that:

• First, treat \(x\) as a function of \(t\), i.e., \(x(t)\).
• Apply the Power Rule, getting \(4x^3\).
• Then multiply by the derivative of \(x\) with respect to \(t\), which is \(\frac{dx}{dt}\).

Hence, the differentiation process for \(x^4\) with respect to \(t\) becomes:
\[\frac{d}{dt} x^4 = 4x^3 \cdot \frac{dx}{dt}\].
A similar process follows for \(y^4\).

Implicit differentiation is a technique used when we have an equation involving two or more variables whose relationship is defined implicitly rather than explicitly. In other words, instead of having one variable isolated on one side of the equation, both variables are mixed together. To differentiate implicitly, follow these steps:

• Differentiate both sides of the equation with respect to the desired variable (in our case, \(t\)).
• Apply the Chain Rule to each variable that is itself a function of the differentiation variable.
• Combine the results and solve for the desired derivative.

In our exercise, we start with the equation \(x^4 + y^4 = 1\). Differentiating both sides with respect to \(t\), and applying the Chain Rule, we get:
\[4x^3 \cdot \frac{dx}{dt} + 4y^3 \cdot \frac{dy}{dt} = 0\].

Finally, isolate \( \frac{dy}{dt} \) to solve for it:
\[\frac{dy}{dt} = -\frac{x^3}{y^3} \cdot \frac{dx}{dt}\].

By doing this, we can find \( \frac{dy}{dt} \) in terms of \(x, y,\) and \( \frac{dx}{dt} \), even though the explicit relationship between \(y\) and \(t\) is hidden.