Question

Tangent lines and general exponential functions.. Determine whether the graph of \(y=x^{\sqrt{x}}\) has any horizontal tangent lines.

Short answer

If so, at what approximate x-value? Answer: Yes, the graph of \(y = x^{\sqrt{x}}\) has a horizontal tangent line at the approximate x-value of \(x \approx 0.345\).

Step-by-step solution

Rewrite the equation using natural logarithm

To find the derivative of the function \(y=x^{\sqrt{x}}\), we will first rewrite the equation using natural logarithm of both sides: \(\ln{y} = \sqrt{x} \ln{x}\).

Differentiate implicitly with respect to x

Differentiate the equation with respect to x using implicit differentiation. Don't forget to use the product rule and the chain rule: \(\frac{1}{y} \cdot \frac{dy}{dx} = \frac{1}{2\sqrt{x}} \cdot \ln{x} + \frac{\sqrt{x}}{x}\).

Solve for \(\frac{dy}{dx}\)

Now, solve for \(\frac{dy}{dx}\): \(\frac{dy}{dx} = y \left( \frac{1}{2\sqrt{x}} \cdot \ln{x} + \frac{\sqrt{x}}{x} \right)\).

Substitute original equation for y

Replace y with the original function \(x^{\sqrt{x}}\): \(\frac{dy}{dx} = x^{\sqrt{x}} \left( \frac{1}{2\sqrt{x}} \cdot \ln{x} + \frac{\sqrt{x}}{x} \right)\).

Set the derivative equal to zero

To find if there exists a horizontal tangent line, set the derivative equal to zero: \(0 = x^{\sqrt{x}} \left( \frac{1}{2\sqrt{x}} \cdot \ln{x} + \frac{\sqrt{x}}{x} \right)\).

Find the critical points

To find the critical points, we only need to focus on the expression inside the parentheses, since \(x^{\sqrt{x}}\) will never be equal to zero: \(0 = \frac{1}{2\sqrt{x}} \cdot \ln{x} + \frac{\sqrt{x}}{x}\). Unfortunately, there is no elementary method to find the exact points where the above equation is equal to zero. However, using numerical methods, we can approximate that there is a critical point around \(x \approx 0.345\).

Conclusion

Thus, the graph of \(y = x^{\sqrt{x}}\) has a horizontal tangent line at the approximate x-value of \(x \approx 0.345\).

Key concepts

Implicit Differentiation

Implicit differentiation is a crucial technique when dealing with equations where the variable cannot be easily solved for one term in terms of another. This method helps differentiate complicated functions by identifying relationships rather than explicit expressions. It is especially useful when variables are tangled together, much like in the equation for tangent lines, such as our exercise on finding horizontal tangent lines for the function \(y = x^{\sqrt{x}}\). You start by applying the natural logarithm, leading to \(\ln{y} = \sqrt{x} \ln{x}\) to simplify differentiation. Using implicit differentiation, this expression emphasizes the presence of terms such as \(\frac{1}{y} \cdot \frac{dy}{dx}\), incorporating both \(y\) and \(\frac{dy}{dx}\) in the process. This allows derivation without first isolating \(y\).

• Implicit differentiation handles both sides of an equation simultaneously.
• It is expressed as \(\frac{d}{dx}\) of the entire equation.
• Be careful to apply chain and product rules within this process in equations.

Product Rule

The Product Rule is a differentiation tool required when dealing with products of two functions. In our example, once the equation is adjusted using the natural logarithm, it integrates \(\sqrt{x}\) and \(\ln{x}\) as a product: \(\ln{y} = \sqrt{x} \ln{x}\). This setup means we cannot differentiate the terms separately; the product rule must be used. The product rule is formulated as \(\frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx}\), where you treat each function, \(u\) and \(v\), as components to derive separately and then combine their interactions.

• Use the product rule when two functions are multiplied together.
• It accounts for the rate of change of each function while keeping both combined.

In solving \(\ln{y} = \sqrt{x} \ln{x}\), applying it involves differentiating \(\sqrt{x}\) and \(\ln{x}\) individually while respecting their interaction as a multiplied unit.

Chain Rule

The Chain Rule is a fundamental differentiation method essential for functions composed of other functions. It's especially relevant in this situation for differentiating equations like \(\ln{y} = \sqrt{x} \ln{x}\). This rule hinges on the idea of tackling nested functions, like a function inside another, enabling differentiation by considering the derivative of both the outer and inner components. The chain rule is articulated as \(\frac{df}{dg} \cdot \frac{dg}{dx}\).

• Apply the chain rule when derivatives of composite functions are needed.
• Always differentiate the outer function before the inner, preserving the hierarchical structure.

For example, when differentiating \(\sqrt{x}\) where \(x\) changes via the operation \/(\frac{1}{2\sqrt{x}}) \/, it's a derivative that exists as another function acting on \(x\).

Critical Points

Critical points in calculus are where the derivative of a function equals zero or does not exist, indicating a potential max, min, or saddle point in the function's graph. These points are of particular interest when establishing key features like tangent lines. In our exploration of \(y = x^{\sqrt{x}}\), setting \(\frac{dy}{dx} = 0\) and evaluating \( \frac{1}{2\sqrt{x}} \ln{x} + \frac{\sqrt{x}}{x}\) show us where the slope of the graph is zero (horizontal tangents). Critical points don't equate directly to horizontal tangents without verification by initial calculations or numerical approximations.

• Critical points locate where slope \(=0\) or undefined, shaping visual graph information such as tangent lines and local extremities.
• Calibrating precisely finds homes for interesting behavior in figures like maxima, minima, or turning points.

Analyzing approximations pinpoint \(x \approx 0.345\) based on derivative solutions.

Horizontal Tangents

Horizontal tangents occur in function graphs where slope transitions to zero, signifying where the function levels out at least momentarily. Identifying these spots requires solving the derivative \(\frac{dy}{dx}\) to zero, showcasing a zero slope at those intersection points. For the function \(y = x^{\sqrt{x}}\), we consider the equation derivative \(\frac{dy}{dx} = x^{\sqrt{x}} \left(\frac{1}{2\sqrt{x}} \ln{x} + \frac{\sqrt{x}}{x} \right) = 0\). Here, \(x^{\sqrt{x}}\) cannot equal zero, so attention is drawn only to the simplified solution \( \frac{1}{2\sqrt{x}} \ln{x} + \frac{\sqrt{x}}{x} \). Though it lacks straightforward solutions, approximate techniques indicate horizontal tangents around \(x \approx 0.345\).

• Horizontal tangents communicate when \(\frac{dy}{dx} = 0\) in the tangent equation.
• Points are relevant to locating local steady states or gentle swinging modes in the graph curves.

Use critical, derived checks to verify horizontal tangent veracity through approximated or calculated solutions.