91Ó°ÊÓ

The Chain Rule is a fundamental tool in calculus used for differentiating composite functions. It allows us to find the derivative of a function with respect to a variable, even if that function is expressed in terms of another variable. In this exercise, the Chain Rule was essential for handling the term \(\tan^{-1}(y)\). This term involves \(y\), which is a function of \(x\), creating a situation where we need the Chain Rule.

• The Chain Rule states that if you have a composite function \(f(g(x))\), then its derivative is \(f'(g(x)) \cdot g'(x)\).
• When differentiating \(\tan^{-1}(y)\), we see \(y\) as an inner function dependent on \(x\). Hence, we apply the Chain Rule, finding first the derivative of \(\tan^{-1}(y)\) with respect to \(y\), resulting in \(\frac{1}{1+y^2}\), and then multiply by \(\frac{dy}{dx}\).
• This enables us to relate changes in \(y\) to changes in \(x\), maintaining the relationship the original equation set forth.

Understanding and correctly applying the Chain Rule is key in dealing with implicit differentiation problems, particularly those involving inverse trigonometric functions like \(\tan^{-1}(y)\).
This ensures that all aspects of the dependency of \(y\) on \(x\) are accounted for in the derivative calculation.

In calculus, the Product Rule is vital for differentiating products of two or more functions. It helps us precisely determine the derivative when one function multiplies another. This becomes essential when terms in equations, such as \(x \tan^{-1}(y)\) in the exercise, involve the product of functions, where each component is dependent on the variable being differentiated.

• The Product Rule states: if you have two functions, \(u(x)\) and \(v(x)\), then the derivative of their product is given by \(u'(x)v(x) + u(x)v'(x)\).
• Using this rule on \(x \tan^{-1}(y)\), we first differentiate \(x\), resulting in \(1\), and multiply by \(\tan^{-1}(y)\).
• Next, we keep \(x\) unchanged and differentiate \(\tan^{-1}(y)\), employing the Chain Rule as \(\frac{1}{1+y^2} \cdot \frac{dy}{dx}\), then add the results.
• This step helps build the complete derivative necessary when differentiating the left side of the given equation.

Effectively applying the Product Rule simplifies complex expressions involving multiplied functions, aiding seamless navigation through implicit differentiation.

Derivatives are the cornerstone of calculus, allowing us to determine the rate of change of one variable with respect to another. In the context of this exercise, derivatives help us to find \(\frac{dy}{dx}\) using implicit differentiation.

• The derivative of a function represents its rate of change and is denoted by \(\frac{d}{dx}\) or as \(f'(x)\).
• Implicit differentiation is used when \(y\), an implicit function of \(x\), is not isolated or directly solvable. Here, treating \(y\) as an implicit function allows differentiation across both sides of equations.
• In this problem, terms such as \(x^3\), \(x \tan^{-1}(y)\), and \(e^y\) are differentiated with respect to \(x\). The derivatives of these terms are calculated using appropriate rules like the Power Rule, Product Rule, and Chain Rule where necessary.

Understanding the concept of derivatives, including its various rules, empowers solving complex problems like the one shown, where composite and implicit functions are differentiated to find \(\frac{dy}{dx}\).
The methodical application of derivatives within this framework clarifies how variables interact and change together, providing deeper insight into implicit relationships.