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When you have a function that is the product of two other functions, you use the product rule to differentiate it. The product rule states that if you have two functions, say \( u(x) \) and \( v(x) \), then the derivative of their product \( u(x)v(x) \) is given by:\[ f'(x) = u'(x)v(x) + u(x)v'(x) \]This means you differentiate one function while keeping the other the same, and then add to it the other way around. Remembering to apply this rule will help you correctly differentiate more complex expressions that involve multiplication.To illustrate, let’s consider the differentiation of the expression in part (a) of the original exercise: \( g(x) = x^2 (3x - 4) \) can be differentiated as follows:Let \( u(x) = x^2 \) and \( v(x) = 3x - 4 \).By the product rule:\[ g'(x) = \frac{d}{dx}(x^2)(3x - 4) + x^2 \frac{d}{dx}(3x - 4) = 2x(3x - 4) + x^2(3) \]\[ = 6x^2 - 8x + 3x^2 = 9x^2 - 8x \]Understanding how to apply the product rule ensures accurate derivative calculations in complex functions.

The quotient rule is used for differentiating a function that is the ratio of two other functions. If we have a function \( f(x) \) that looks like \( \frac{u(x)}{v(x)} \), then according to the quotient rule, its derivative is:\[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} \]So, you differentiate the numerator and the denominator separately, multiply the numerator's derivative by the denominator, subtract the product of the numerator and the derivative of the denominator, and then divide it by the square of the original denominator.Let's apply the quotient rule to differentiate \( j(x) = \frac{x^2 - 3}{x} \), part (d) in the original exercise. First, rewrite it as:\( j(x) = x - \frac{3}{x} \).Then differentiate each part:\[ j'(x) = \frac{d}{dx}(x) - \frac{d}{dx}(\frac{3}{x}) \]\[ = 1 - (- \frac{3}{x^2}) \]\[ = 1 + \frac{3}{x^2} \]Finally, rewrite it in a single fraction if needed:\[ = \frac{x^2 + 3}{x^2} \]Learning the quotient rule ensures you can handle more intricate fraction-based functions confidently.

Polynomial differentiation refers to the process of finding the derivative of polynomial functions. These are expressions consisting of variables raised to powers and coefficients. The power rule is straightforward: for a term \( ax^n \), the derivative is given by \( anx^{(n-1)} \).For instance, consider the function \( k(x) = \frac{x - 4x^2}{x^3} \) from part (e) of the original exercise. First, rewrite it as:\( k(x) = \frac{1}{x^2} - 4\frac{1}{x} \).Then differentiate each term individually:\[ k'(x) = -2x^{-3} + 4x^{-2} \]Convert the power notation to return to fraction form:\[ = -\frac{2}{x^3} + \frac{4}{x^2} \]Knowing the basic rules of polynomial differentiation, such as the power rule, is essential for accurately and quickly handling a variety of calculus problems.

Simplification of expressions is a crucial step in solving and differentiating complex functions. Simplifying before differentiation can save you time and reduce errors, as seen in the original exercise.For example, consider part (f): \( l(x) = \frac{x^2 - 3x + 5}{x^2} \).First, rewrite it to simplify the expression:\[ l(x) = 1 - \frac{3x}{x^2} + \frac{5}{x^2} \]Rewrite again for clarity:\[ = 1 - \frac{3}{x} + \frac{5}{x^2} \]Now it’s easier to differentiate:\[ l'(x) = 0 - (-\frac{3}{x^2}) + 2 \frac{5}{x^3} = \frac{3}{x^2} - \frac{10}{x^3} \]Simplifying expressions before differentiating ensures you handle fewer terms and smaller powers, which can make your calculations much easier and less prone to mistakes.

Calculating derivatives accurately requires knowing the basic rules and methods of differentiation. These include the power rule, product rule, quotient rule, and chain rule.For instance, let's revisit part (c) from the original exercise: \( i(x) = (x+1)(x-6) \).Using the product rule, we differentiate as follows:Let \( u(x) = x + 1 \) and \( v(x) = x - 6 \).According to the product rule:\[ i'(x) = (1)(x-6) + (x+1)(1) \]\[ = x - 6 + x + 1 \]\[ = 2x - 5 \]Hence, comprehensively applying derivative rules and systematically handling each term within your function ensures precise and accurate derivative calculations.