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The chain rule is a fundamental concept in calculus, particularly when dealing with composite functions. It helps us find the derivative of a function that is nested inside another function. For the equation \( y = \arctan\left(\frac{x-1}{x+1}\right) \), the chain rule is employed since \( y \) involves an arctan of a fraction involving \( x \).

To use the chain rule, we first need to differentiate the outer function, which in this case is \( \arctan(u) \). The derivative of \( \arctan(u) \) is \( \frac{1}{1+u^2} \).

• Differentiating the outer function: \( \frac{d}{du} \arctan(u) = \frac{1}{1+u^2} \).
• Differentiating the inner function \( u = \frac{x-1}{x+1} \) by applying the quotient rule.

Once both derivatives are found, they are multiplied together to give \( \frac{dy}{dx} = \frac{1}{1+u^2} \cdot \frac{du}{dx} \), which is a perfect demonstration of the chain rule in action.

Inverse trigonometric functions, such as the arctan, allow us to find the angles given a trigonometric value. In calculus, their derivatives are crucial when we deal with equations involving angles.

For example, for \( y = \arctan\left(u\right) \), we differentiate using the formula \( \frac{dy}{du} = \frac{1}{1+u^2} \). This expression dictates how quickly the angle changes concerning \( u \).

• This derivative is particularly useful because it simplifies the process, especially when coupled with the chain rule.
• It also shows us the behavior of the inverse trigonometric function over different values of \( u \).

Recognizing the forms and derivatives of these functions is key when applying them in various calculus problems.

When dealing with derivatives of functions that are fractions, the quotient rule is employed. It assists in differentiating complex fractional expressions like \( u(x) = \frac{x-1}{x+1} \).

The quotient rule states that if you have two functions \( f(x) \) and \( g(x) \), the derivative of their quotient \( \frac{f(x)}{g(x)} \) is given by:

• \( \frac{d}{dx} \left[ \frac{f(x)}{g(x)} \right] = \frac{g(x)f'(x) - f(x)g'(x)}{(g(x))^2} \).

In this exercise, for \( u(x) = \frac{x-1}{x+1} \), \( f(x) = x-1 \) and \( g(x) = x+1 \):

• The derivative is \( \frac{(x+1)(1) - (x-1)(1)}{(x+1)^2} \).
• Upon simplification, this becomes \( \frac{2}{(x+1)^2} \).

Applying the quotient rule helps in accurately finding derivatives of ratios, which is a common requirement in calculus tasks.

Simplification in calculus is an essential step that transforms complicated expressions into a more manageable form. It's crucial in our example when we derive the expression \( \frac{dy}{dx} \) in order to reach a clean and clear final answer.

After differentiating \( y = \arctan\left(\frac{x-1}{x+1}\right) \), simplification takes place to deal with complex fractions and powers:

• Simplifying \( 1 + \left( \frac{x-1}{x+1} \right)^2 \) involves expanding and combining like terms using algebraic identities.
• Cancelling common factors assists in reducing the expression \( \frac{2}{2x^2 + 6x + 2} \) to a simpler form.

This process not only makes calculations easier but also provides insight into the behavior of the function and its derivative. Simplification is an art in calculus that demands understanding both algebraic manipulation and the logic of calculus.