91Ó°ÊÓ

Differentiation is a fundamental concept in calculus that helps us understand how a function changes. The process involves finding a derivative, which represents the rate of change of a function with respect to a variable. For the equation given in this problem, the function is expressed as \(y = x - \sqrt{x}\). To find out how this function changes with respect to \(x\), we perform differentiation.

Here's what happens during differentiation:

• The derivative of \(x\) with respect to \(x\) is simply 1, because \(x\) increases at a constant rate.
• The term \(-\sqrt{x}\) is rewritten as \(-x^{\frac{1}{2}}\) to apply the power rule, which involves bringing down the exponent and subtracting one from it.
• Using the power rule, the derivative becomes \(-\frac{1}{2}x^{-\frac{1}{2}}\).

By combining these derivatives, the overall derivative of the function \(y\) becomes \(y' = 1 - \frac{1}{2}x^{-\frac{1}{2}}\). This result tells us the instantaneous rate of change or the slope of the curve for any given \(x\).

A tangent line to a curve at a given point is a straight line that touches the curve at that point without crossing it. We often use tangent lines to approximate values of functions or to find the slope at a point. The slope of this tangent line is given by the derivative, which we calculated earlier.

In the problem, we are interested in finding points where the tangent line is horizontal. When a line is horizontal, its slope is zero. Thus, we set the derivative \(y' = 1 - \frac{1}{2}x^{-\frac{1}{2}}\) equal to zero to find these points. This means solving for \(x\) when:

• \(1 - \frac{1}{2}x^{-\frac{1}{2}} = 0\)

These x-values will tell us where on the curve the tangent line is perfectly flat, indicating no net increase or decrease at that point.

The slope of a curve at a particular point refers to how steep or flat the curve is at that point. Calculus allows us to dynamically calculate this slope through differentiation, producing the derivative function \(y'\). By evaluating this derivative at any point \(x\), we determine the slope there.

In this context, the derivative \(y' = 1 - \frac{1}{2}x^{-\frac{1}{2}}\) denotes the slope of the original function \(y = x - \sqrt{x}\) at any x-value. To find where the slope is zero (a horizontal line), we set \(y'\) equal to zero.

• Solve \(1 - \frac{1}{2}x^{-\frac{1}{2}} = 0\)
• After solving, we found that \(x = \frac{1}{4}\) is the point where the slope is zero.

This means that at \(x = \frac{1}{4}\), the tangent line to the curve is horizontal, with the curve neither rising nor falling at that location.