91Ó°ÊÓ

The product rule is a fundamental concept in calculus. It is used when differentiating a product of two functions. In our exercise, we begin with the equation\[xy = \cot(xy)\]. When differentiating the left-hand side, we need to apply the product rule because we have two functions, \(x\) and \(y\), multiplied together. The product rule tells us that the derivative of \(uv\) (where \(u\) and \(v\) are functions of \(x\)) is \(u'v + uv'\).

To apply the product rule here: - Treat \(x\) as one function and \(y\) as another. Since \(x\) is simply \(x\), its derivative is 1.- \(y\), being a function of \(x\), will have the derivative \(\frac{dy}{dx}\) when implicitly derived.

Therefore, differentiating \(xy\) using the product rule gives us \(x\frac{dy}{dx} + y\). This step is crucial to understanding how changes in \(x\) influence \(y\) within this context.

The chain rule is another essential element of calculus used for differentiating composite functions. When you have a function inside another function, the chain rule is applied to differentiate them efficiently. In our problem, the equation \(xy = \cot(xy)\) requires using the chain rule on the right-hand side.

The cotangent function, \(\cot(xy)\), is a composite function: we have \(\cot\), a trigonometric function, with \(xy\) as its argument that itself is a function of \(x\). Therefore, by the chain rule, the derivative of \(\cot(u)\) with respect to \(u\) is \(-\csc^2(u)\). We then multiply this by the derivative of \(u=xy\) with respect to \(x\), which is \(x\frac{dy}{dx} + y\) as derived using the product rule.

• The chain rule requires applying the outer function's derivative first: \(-\csc^2(xy)\).
• Then multiply it by the inner function's derivative: \((x \frac{dy}{dx} + y)\).

This results in \(-\csc^2(xy) \cdot (x \frac{dy}{dx} + y)\), showing how complex functions can be tackled step by step.

Trigonometric functions are frequently encountered in calculus. They often require application of rules such as the product and chain rules. In this example, understanding \(\cot\) and its derivative is crucial to the solution.

- \(\cot(x)\), the cotangent of \(x\), is defined as the ratio of \(\cos(x)\) to \(\sin(x)\).- Its derivative, which appears in implicit differentiation, is \(-\csc^2(x)\). This is because \(\csc(x) = 1/\sin(x)\), leading to the identity \(\csc^2(x) = (1/\sin(x))^2\).

When derivatives involve trigonometric functions, they often appear in implicit differentiation due to their cyclical and repeating nature in functions over different distances or angles. Recognizing and using these derivatives correctly allows us to handle the equation effectively, leading to simplification and solution. Trigonometric identities and derivatives are keys that unlock many complicated calculus problems, like the one in this exercise.