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The chain rule is a fundamental concept in calculus used when dealing with composite functions. A composite function is a function made up of other functions. In simpler terms, it's a function within a function. For example, in this exercise, we have a function inside the natural logarithm:

• Outer function: \( y = \ln(u) \)
• Inner function: \( u = x^{5/2} \)

To find the derivative of such a composite, you first take the derivative of the outer function with respect to the inner function and then multiply it by the derivative of the inner function with respect to \( x \). This principle can be formally expressed as: If \( y = f(g(x)) \), then \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \). Just think of it as peeling layers of an onion, addressing the outer layer first and moving inward!

A derivative represents the rate at which a function is changing at any given point, and it is a cornerstone of calculus. It can be thought of as the slope of the function at a certain point. In this exercise, the function is \( y = \ln(x^{5/2}) \).To find the derivative, we apply the chain rule. Here, the derivative of the outer function, \( \ln(u) \), is \( \frac{1}{u} \), and the derivative of the inner function, \( x^{5/2} \), is \( \frac{5}{2}x^{3/2} \).By multiplying these derivatives together, according to the chain rule, we obtain \[ y' = \frac{1}{x^{5/2}} \times \frac{5}{2}x^{3/2} = \frac{5}{2x} \] This result gives you the formula to find the slope at any point \( x \) along the curve defined by our original function.

The slope of the tangent line to a curve at a given point is precisely the value of the derivative at that point. Tangently, this can be imagined as the steepness of the curve at a specific location. When you picture a graph, the tangent line is the straight line that just "kisses" the curve at one point without crossing it. It gives a perfect picture of the curve's immediate direction and steepness.In our exercise, with the function \( y = \ln(x^{5/2}) \), we identify the point of interest as \((1,0)\).At this point, we substitute \( x = 1 \) into the derivative we calculated, \( y' = \frac{5}{2x} \). So the slope of the tangent line is:\[ y'(1) = \frac{5}{2 \times 1} = \frac{5}{2} \]The significance of this slope, \( \frac{5}{2} \), is that it tells you exactly how steep the graph is at the point (1,0). It aids in drawing the tangent line accurately if needed.

Function evaluation involves computing the value of a function at a specific point. In calculus, after finding a derivative, evaluating this derivative at a particular point helps us solve problems concerning rates of change or give insight into the behavior of the function at that point.In the exercise, after determining the expression for the derivative, \( y' = \frac{5}{2x} \), we performed function evaluation at the point \( x = 1 \).This is how it looked:

• Plug \( x=1 \) into the derivative: \( y'(1) = \frac{5}{2 \times 1} \)
• Compute the result: \( \frac{5}{2} \)

Evaluating a derivative at a specific point like this allows us to make precise observations about the function's behavior at that point, such as understanding the rate at which the function is increasing or decreasing.