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When you need to differentiate a function that is a product of two other functions, the product rule is your go-to technique. It helps us find the derivative of a function like \(f(x) = u(x) \cdot v(x)\). The formula for the product rule is given by:

• \( (u \cdot v)' = u' \cdot v + u \cdot v' \)

This means that you differentiate each function individually, then combine them according to the formula. In our example, we have two functions to differentiate:

• \(u = x^2\) with the derivative \( u' = 2x \)
• \(v = \sec^{-1}(5x)\), which we'll explore further below.

So the product rule helps us split the work into manageable steps, ensuring nothing is left out. Remember, always differentiate separately and then apply the product rule formula.

Inverse trigonometric functions, like \(\sec^{-1}(x)\), have specific rules for differentiation. These functions "undo" the trigonometric functions, which can be quite useful in calculus. For \(\sec^{-1}(u)\), the derivative is given by:

• \( \frac{d}{dx}[\sec^{-1}(u)] = \frac{1}{|u|\sqrt{u^2 - 1}} \cdot \frac{du}{dx} \)

The key part is the chain rule element, as you multiply by \( \frac{du}{dx} \), where \(u\) is typically a function of \(x\). In our context, \(u = 5x\), and thus \(\frac{du}{dx} = 5\).After plugging in this derivative back into our product rule, it's simplified further:

• The derivative becomes \( v' = \frac{1}{x\sqrt{25x^2 - 1}}\) for positive \(x\).

Understanding derivatives for these inverse trigonometric functions allows you to tackle complex expressions in calculus with more confidence.

The chain rule is a powerful tool in calculus for differentiating composite functions, essentially functions within other functions. It states that if you have a function \( f(g(x)) \), the derivative is:

• \( \frac{d}{dx}[ f(g(x))] = f'(g(x)) \cdot g'(x) \)

In simple terms, differentiate the outer function, then multiply by the derivative of the inner function.Let's see how this works with our inverse trigonometric function \(\sec^{-1}(5x)\). In this scenario, the outer function \(f\) is \(\sec^{-1}(u)\), and the inner function \(g(x)\) is \(u = 5x\). Differentiating the outer function while treating the inner function as a variable gives us part of the derivative, then multiplying by \(\frac{du}{dx} = 5\) completes it.This is crucial when combining the chain rule with the product rule. Applying these rules step-by-step ensures clarity and accuracy when calculating derivatives.