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Derivatives are fundamental in calculus and are used to understand how a function changes as its input changes. In implicit differentiation, we find the derivative of a function that is not isolated for one variable. For example, consider the equation \( \frac{1}{x} + \frac{1}{y} = 1 \). We can differentiate this equation with respect to \( x \), even though \( y \) is implicitly defined within the equation.

• Start by differentiating each term: \( \frac{1}{x} \) differentiates to \( -\frac{1}{x^2} \).
• For \( \frac{1}{y} \), apply the chain rule, resulting in \(-\frac{1}{y^2} y'\).

Using the implicit differentiation method helps reveal relationships between variables even when they are not explicitly presented. Understanding derivatives in this context is key for advanced mathematical analysis.

The chain rule is a method used in calculus for finding the derivative of composite functions. It is particularly useful in implicit differentiation. When dealing with a composite function, such as \( \left(\frac{1}{y}\right) \) where \( y \) is a function of \( x \), the chain rule helps to break down the derivatives.Here’s how it works:- If \( y = g(x) \), then for \( \, \frac{d}{dx}(\frac{1}{y}) = - \frac{1}{y^2} \cdot \frac{dy}{dx} \).- This formula breaks down the differentiation into more manageable parts by considering the derivative of the outer layer and the derivative of the inner function separately.By mastering the chain rule, you can tackle more complex derivative problems with confidence, such as those encountered in implicit differentiation.

The quotient rule is applied when differentiating a function that is the quotient of two other functions. It’s essential when finding the derivative of expressions like \( y = \frac{x}{x-1} \). According to the quotient rule, if \( u(x) \) and \( v(x) \) are differentiable functions:\[ f(x) = \frac{u}{v} \Rightarrow f'(x) = \frac{u'v - uv'}{v^2} \]For \( y = \frac{x}{x-1} \):- Let \( u = x \) (so \( u' = 1 \)) and \( v = x-1 \) (so \( v' = 1 \))- The derivative is calculated as: \[ y' = \frac{(x-1)(1) - x(1)}{(x-1)^2} = \frac{-1}{(x-1)^2} \]Using the quotient rule helps ensure accurate differentiation of complex rational functions, an invaluable tool in calculus.

A consistency check is an important step in validating the results derived from different methods of differentiation. After determining derivatives implicitly and explicitly, ensure both results match to verify accuracy. This involves substituting the explicit solution back into the implicitly derived expression.For the exercise:- From implicit differentiation: \( y' = \frac{y^2}{x^2} \).- By substituting \( y = \frac{x}{x-1} \), we get: \( y^2 = \left( \frac{x}{x-1} \right)^2 \).- Substitute back into \( y' \):\[ y' = \frac{\left( \frac{x}{x-1} \right)^2}{x^2} = \frac{1}{(x-1)^2} \]A successful consistency check in this situation confirms that both methods lead to the same derivative, ensuring reliability and correctness of the solution. It's a vital part of problem-solving that reduces errors and builds confidence in mathematical operations.