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If \(g\) is a differentiable function, find an expression for the deriv- ative of each of the following functions. $$\text { (a) } y=x g(x) \quad \text { (b) } y=\frac{x}{g(x)} \quad \text { (c) } y=\frac{g(x)}{x}$$

Short Answer

Expert verified
For (a): \( xg'(x) + g(x) \), for (b): \( \frac{g(x) - xg'(x)}{(g(x))^2} \), and for (c): \( \frac{xg'(x) - g(x)}{x^2} \).

Step by step solution

01

Use the Product Rule for Part (a)

To find the derivative of the function \( y = xg(x) \), we use the product rule. The product rule states that the derivative of a product of two functions \( u(x) \) and \( v(x) \) is given by: \((uv)' = u'v + uv'\). Here, \( u(x) = x \) and \( v(x) = g(x) \).\\[\frac{dy}{dx} = xg'(x) + g(x)\].
02

Use the Quotient Rule for Part (b)

To differentiate \( y = \frac{x}{g(x)} \), we apply the quotient rule. The quotient rule states: if \( u(x) \) and \( v(x) \) are functions, then their quotient's derivative \(\left(\frac{u}{v}\right)'\) is \( \frac{u'v - uv'}{v^2} \). Here, \( u(x) = x \) and \( v(x) = g(x) \), thus, \( u'(x)=1 \).\\[\frac{dy}{dx} = \frac{g(x) - xg'(x)}{(g(x))^2}\].
03

Use the Quotient Rule for Part (c)

For \( y = \frac{g(x)}{x} \), use the quotient rule again. Here, \( u(x) = g(x) \) and \( v(x) = x \), so \( u'(x) = g'(x) \) and \( v'(x) = 1 \). Apply the quotient rule: \( \frac{u'v - uv'}{v^2} \).\\[\frac{dy}{dx} = \frac{xg'(x) - g(x)}{x^2}\].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Product Rule
When you have a function that is a product of two simpler functions, the product rule comes in handy during differentiation. Let’s say you have two differentiable functions, \( u(x) \) and \( v(x) \). The product rule provides a formula to find the derivative of their product: \((uv)' = u'v + uv'\). This essentially means that to differentiate the product of \( u \) and \( v \), you differentiate \( u \) and multiply it by \( v \), then differentiate \( v \) and multiply it by \( u \), and finally add the results together. This rule is particularly useful when dealing with real-world problems involving rates of change, where multiple factors are changing at the same time.
For example, consider the function \( y = xg(x) \). Here, \( u(x) = x \) and \( v(x) = g(x) \). To find the derivative \( \frac{dy}{dx} \), you use the product rule as follows:
\[\frac{dy}{dx} = xg'(x) + g(x).\]
This way, you account for the changes in both \( x \) and \( g(x) \). The result is a new function that represents the rate of change of your original function.
Quotient Rule
The quotient rule is essential when you’re working with fractions or ratios of two functions. Similar to the product rule, but it applies to the division of functions. If you have functions \( u(x) \) and \( v(x) \), the quotient rule states that the derivative of their ratio is given by: \( \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \). This rule helps you deal with complex functions where one function is divided by another. Always remember: differentiate the numerator and the denominator separately, multiply them cross-wise, then subtract and divide by the square of the denominator.
For instance, with the function \( y = \frac{x}{g(x)} \), the quotient rule applies. Here, \( u(x) = x \) and \( v(x) = g(x) \). Using the quotient rule, the derivative becomes:
\[\frac{dy}{dx} = \frac{g(x) - xg'(x)}{(g(x))^2}.\]
This formula helps you understand the behavior of the function \( y \) by evaluating how quickly its value changes relative to your choice of \( x \) and \( g(x) \).
Similarly, for \( y = \frac{g(x)}{x} \), where \( u(x) = g(x) \) and \( v(x) = x \), apply the quotient rule again:
\[\frac{dy}{dx} = \frac{xg'(x) - g(x)}{x^2}.\]This method is consistently used in physics and economics to analyze rates.
Derivative of a Function
Understanding the derivative of a function is crucial in calculus. A derivative represents the rate at which a function changes at any given point. It gives us the function's slope at a particular point, thus providing insights into the function's behavior, such as identifying maximum and minimum values or points of inflection.
Finding the derivative involves applying rules like the product or quotient rule, depending on the complexity and type of function you are dealing with.
  • The product rule is helpful when the function is a product of two simpler functions.
  • The quotient rule is used when the function is the ratio of two functions.
These rules help decompose complex problems into manageable parts, making it possible to calculate derivatives efficiently. Knowing how to apply these differentiation techniques helps in various fields such as physics, engineering, and economics, where understanding change is crucial.

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