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Derivatives are a fundamental concept in calculus and represent the rate of change of a function with respect to a variable. In simpler terms, a derivative tells us how a function's output value changes as its input value changes.

The derivative of a function is often denoted by a prime symbol, such as \( f'(x) \) for the function \( f(x) \). To compute derivatives, basic rules such as the power rule, product rule, quotient rule, and chain rule are used. Each rule has specific applications depending on the form of the function.

For example, in the function \( j(x) = 5e^{-x} + \ln x \), the derivative \( j'(x) \) is found by differentiating each term individually. The term \( 5e^{-x} \) becomes \(-5e^{-x}\) with the chain rule, and \( \ln x \) becomes \( \frac{1}{x} \) because the derivative of the natural logarithm is \( \frac{1}{x} \). Ultimately, the derivative is expressed as \( j'(x) = -5e^{-x} + \frac{1}{x} \).

Critical points are where a function's derivative is zero or undefined. These points are important because they can signal where a function is about to change direction. Knowing these points helps in analyzing and graphing functions, as they typically indicate potential local maxima or minima.

To find critical points for the function \( j(x) = 5e^{-x} + \ln x \), we set the derivative \( j'(x) = -5e^{-x} + \frac{1}{x} \) to zero. Solving \( -5e^{-x} + \frac{1}{x} = 0 \) gives the critical equation \( x = \frac{1}{5} e^x \). This type of equation may not have a straightforward algebraic solution, so numerical and graphical methods are typically used for finding values like \( x \approx 0.474 \).

These calculated points are then used for further analysis to determine if they are maxima, minima, or neither.

Relative extrema refer to the local maximum and minimum of a function. These are points where the function reaches a peak (maximum) or valley (minimum) within a certain range.

To determine relative extrema for the function \( j(x) \), we look at its critical points located previously. Once we have the critical point, around \( x \approx 0.474 \), we use the second derivative test to identify the nature of this point.

Substituting back into the second derivative \( j''(x) = 5e^{-x} - \frac{1}{x^2} \) helps in asserting whether the function is concave up or down at this point. If \( j''(x) > 0 \), the function has a local minimum, whereas \( j''(x) < 0 \) would indicate a local maximum. For \( x \approx 0.474 \), the second derivative is positive, confirming a local minimum.

The chain rule is an essential tool in calculus for differentiating composite functions. It provides a formula to compute the derivative of a function that consists of another function inside it (a composition of functions).

The chain rule states that if you have a composite function \( h(x) = f(g(x)) \), the derivative \( h'(x) \) is given by \( h'(x) = f'(g(x)) \cdot g'(x) \). This means you first find the derivative of the outer function evaluated at the inner function, and then multiply it by the derivative of the inner function.

In the function \( j(x) = 5e^{-x} + \ln x \), we applied the chain rule to the term \( 5e^{-x} \). For this, \( e^{-x} \) is treated as a composite of \( e^u \) with \( u = -x \). The derivative of \( e^u \) with respect to \( u \) is \( e^u \), and the derivative of \( -x \) with respect to \( x \) is \(-1\). Combining these gives the derivative \( -5e^{-x} \), demonstrating the chain rule's application effectively.