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When working with calculus, especially when finding the derivative of products of functions, the product rule becomes a vital tool. If you have a function that can be expressed as a product of two functions, say \( u(x) \) and \( v(x) \), then the derivative of this product is not simply the product of their derivatives. Instead, we use the product rule. This rule states that the derivative of a product is the derivative of the first function times the second function, plus the first function times the derivative of the second function. The formula is written as:

• \((uv)' = u'v + uv' \)

This means you'll need to find the derivative of each function separately before applying the formula. In the context of the original exercise, where \( u(x) = 2.5(0.9^x) \) and \( v(x) = \ln(x) \), apply the product rule to effectively find \( f'(x) \). Don't forget, each function might require different derivative techniques like the chain rule, as shown in the solution.

The chain rule is another crucial differentiation technique, particularly for composite functions. A composite function is essentially a function inside another function. To differentiate such a function, you'll need the chain rule, which helps to "unpack" these layers.In the original problem, \( u(x) = 2.5(0.9^x) \) contains a composite structure because of the exponent \( 0.9^x \). To differentiate \( 0.9^x \), you apply the chain rule, which involves finding the derivative of the outer function and multiplying it by the derivative of the inner function. The rule can be summarized as:

• If \( y = g(h(x)) \), then \( \frac{dy}{dx} = g'(h(x)) \cdot h'(x) \)

For our specific function, we identify \( 0.9^x \) as the inner function. Its derivative is \( \ln(0.9) \. 0.9^x \), meaning you recognize the exponential term and multiply by the natural logarithm of the base, \( \ln(0.9) \). Combining this with the constant multiplier 2.5 completes the differentiation of \( u(x) \) using the chain rule.

Differentiation techniques are varied and powerful tools in calculus, enabling us to find the rate of change of functions. Understanding when and how to use different rules is key to solving derivative problems correctly and efficiently.Here are some essential differentiation techniques:

• Power rule: Useful when differentiating terms like \( x^n \). Simply bring down the exponent and reduce it by one: \( nx^{n-1} \).
• Product rule: As explained, used when dealing with products of functions. Essential for terms like \( u(x) â‹… v(x) \).
• Chain rule: Critical for composite functions, allowing the differentiation of nested functions.
• Quotient rule: Used for functions of the form \( \frac{u(x)}{v(x)} \). This handles divisions and is structured similarly to the product rule but involves subtraction.
• Logarithmic differentiation: Particularly handy for complex equations involving logarithms, where you initially take the natural log to simplify the function.

Applying these techniques accurately will enable you to handle various functions and attain their derivatives. Each method has its specific application, ensuring efficient and correct problem-solving. For the original exercise, recognition of where to apply each technique was crucial in finding the correct solution.