91Ó°ÊÓ

Understanding Theorem 15.9 is key to applying implicit differentiation effectively. Implicit differentiation involves finding derivatives of functions implicitly defined by an equation. The theorem essentially allows us to differentiate both sides of an equation concerning one variable, usually when one variable is defined in terms of the other. This process is essential when a function is not explicitly solved for one variable.

When using Theorem 15.9 in the context of \[ y \ln (x^2 + y^2 + 4) = 3 \]we consider \( y \) as a differentiable function of \( x \). By applying the theorem, we address the presence of \( y \) implicitly embedded in the logarithm argument, making differentiation feasible through implicit means.

Remember, the theorem is not only about differentiating both sides but doing so while recognizing each variable's dependence on others in the equation.

The product rule is a fundamental tool in calculus for finding the derivative of a product of two functions. It describes how to differentiate expressions where two variables are multiplied together and need to be treated differently from a simple summation.

The product rule states:\[\frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x)\]In this exercise, we applied the product rule by setting \( u(x) = y \) and \( v(x) = \ln(x^2 + y^2 + 4) \). So, we found the derivative of each individually and then employed the rule:

• Differentiate \( y \) with respect to \( x \), denoted as \( \frac{dy}{dx} \).
• Compute the derivative of \( \ln(x^2 + y^2 + 4) \) using chain rule.

Lastly, substitute these derivatives back into the rule to achieve the left-hand side differentiation, vital for solving implicit equations.

In mathematics, a function is called differentiable if it has a derivative, meaning it can be differentiated at every point in its domain. This guarantee of differentiability is crucial in calculus as it allows us to apply standard rules and theorems, such as those we use in implicit differentiation.

In the given exercise: \[ y \ln (x^2 + y^2 + 4) = 3 \]we assume \( y \) is a differentiable function of \( x \). This assumption means we can perform calculus operations, like derivative computations, reliably. Differentiability guarantees that \( y \) behaves smoothly, without any abrupt changes or breaks in the slope of the function.

Differentiable functions typically arise from continuous functions. Continuity ensures the function does not have gaps or jumps. Therefore, verifying differentiability is the first step before any derivative computations are carried out, ensuring the solution's correctness.

Derivative computation is the process of finding the derivative or rate of change of a function with respect to a variable. In implicit differentiation, this involves taking derivatives of both sides of the equation concerning that variable and handling any mixed derivatives carefully.

From our example, differentiating: \[ y \ln(x^2 + y^2 + 4) = 3 \]requires finding:

• The derivative of \( y \ln(x^2 + y^2 + 4) \), which involves both product rule and chain rule.
• The constant derivative \( \frac{d}{dx}(3) = 0 \), as constants have a zero rate of change.

After computing these derivatives separately, they're compiled into an equation that is solved for \( \frac{dy}{dx} \). The steps of distributing, isolating, and solving are core to evaluating the slope of \( y \) in the original context, despite the complexity of its embedded nature within other functions.

Understanding these computation techniques is crucial for mastering complex differentiation problems in calculus.