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The chain rule is one of the essential tools in calculus for taking derivatives of composite functions. It helps us differentiate a function that is wrapped inside another function. For example, in the function \( \tan(xy) \), there is an outer function \( \tan(u) \) where \( u = xy \).

• First, identify the inner and outer functions.
• Differentiate the outer function: the derivative of \( \tan(u) \) is \( \sec^2(u) \).


• Differentiate the inner function \( xy \) (using the product rule in this case, which we will cover next).

The key to applying the chain rule is to multiply the derivative of the outer function by the derivative of the inner function. In the exercise provided, after applying the chain rule, we obtain an expression involving \( \sec^2(xy) \) and the derivative of \( xy \). This showcases the power of the chain rule in simplifying and systematically breaking down complex differentiation tasks.

The product rule is crucial when dealing with the derivative of two functions multiplied together. It states that if you have two functions, \( u \) and \( v \), the derivative of their product \( uv \) is given by \( u'v + uv' \).

• Identify the two functions, \( u = x \) and \( v = y \), in the product \( xy \).
• Differentiate each separately: \( u' = 1 \) and \( v' = \frac{dy}{dx} \) (as \( y \) is a function of \( x \)).


• Apply the product rule: \( 1 \cdot y + x \cdot \frac{dy}{dx} \).

The product rule allows us to handle scenarios where two functions are intertwined through multiplication, each possibly dependent on another variable, like \( x \). In this implicit differentiation problem, using the product rule within the chain rule extends our ability to solve for derivatives even in multi-variable contexts.

Implicit differentiation is a powerful technique used when a function \( y \) is not isolated on one side of the equation but instead, is mingled with \( x \). When we cannot easily solve for \( y \) explicitly, implicit differentiation allows us to differentiate both sides of the equation with respect to \( x \), treating \( y \) as a function of \( x \).

• Differentiate each term of the equation with respect to \( x \).
• Involve \( \frac{dy}{dx} \) directly wherever \( y \) appears.


• Solve for \( \frac{dy}{dx} \) in the resulting equation.

In this exercise, implicit differentiation enables us to find \( \frac{dy}{dx} \) where the equation consists of \( x \) and \( \tan(xy) \), recognizing that \( y \) implicitly depends on \( x \). By using both chain and product rules, implicit differentiation is adeptly applied to differentiate composite and product functions within the implicit context.