91Ó°ÊÓ

Differentiation is a fundamental tool in calculus. It allows us to determine how a function changes with respect to its variables. For the function \(y = x \text{ln} x\), we use differentiation to find the rate of change or the slope of the curve at any point. The first derivative, \(\frac{dy}{dx}\), gives us this slope. Using the product rule, we find \(\frac{dy}{dx} = \text{ln} x + 1\). Setting the derivative to zero helps us identify critical points, which are points where the slope of the curve is zero. These are potential turning points where the function may change from increasing to decreasing, or vice versa. In this exercise, solving \(\text{ln} x + 1 = 0\) gives the critical point \(x = \frac{1}{e}\). This is an application of logarithmic differentiation, which is useful for functions involving logarithms.

Turning points are crucial features of curves where the function changes direction, either from increasing to decreasing (a maximum) or from decreasing to increasing (a minimum). For the given function \(y = x \text{ln} x\), once we find the critical point at \(x = \frac{1}{e}\), we determine the corresponding \(y\)-coordinate by substituting this \(x\) back into the original function: \(y = \frac{1}{e} \text{ln}(\frac{1}{e}) = -\frac{1}{e}\). The turning point’s coordinates are \(\bigg(\frac{1}{e}, -\frac{1}{e}\bigg)\). To classify this turning point, we use the second derivative, \(\frac{d^2y}{dx^2}\). If it’s positive, the turning point is a minimum; if negative, it is a maximum. Here, \(\frac{d^2y}{dx^2} = \frac{1}{x}\), which evaluates to \(e\) at \(x = \frac{1}{e}\), indicating a minimum. Understanding this process helps in analyzing and sketching the behavior of functions.

Parametric equations define curves where both \(x\) and \(y\) are expressed as functions of a separate parameter, \(t\). In the exercise, the curves are given by \(x = t - \frac{1}{t}\) and \(y = t + \frac{1}{t}\). Each value of \(t\) gives a corresponding \((x, y)\) point. To find the gradient of the curve without combining \(x\) and \(y\) into one function, we differentiate both equations with respect to \(t\): \(\frac{dx}{dt} = 1 + \frac{1}{t^2}\) and \(\frac{dy}{dt} = 1 - \frac{1}{t^2}\). The gradient \(\frac{dy}{dx}\) is then found as \(\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\). Set the gradient to zero gives us the points where the slope is zero. For \(t = 1\) or \(t = -1\), the equation yields points (0,2) and (0,-2), respectively. Understanding parameterization provides a powerful way to handle complex curves.

Tangent lines touch a curve at just one point and have the same slope as the curve at that point. To find the equation of a tangent line, you need the point of tangency and the slope of the curve at that point. From the parametric equations given in the exercise, we find the coordinates at \(t = 2\) by substituting it into \(x = t - \frac{1}{t}\) and \(y = t + \frac{1}{t}\). This gives \((\frac{3}{2}, \frac{5}{2})\). To determine the slope, we evaluate \(\frac{dy}{dx}\) at \(t = 2\): \(\frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{\frac{3}{4}}{\frac{5}{4}} = \frac{3}{5}\). Therefore, the equation of the tangent line is \(y - \frac{5}{2} = \frac{3}{5} (x - \frac{3}{2})\). Tangent lines are useful in approximating curves locally and understanding their behavior.