We are tasked to calculate the gradient of the tangents to the curve at $x = 1$, where the curve is defined as:

$y = \frac{\sqrt{2x - 1}}{x}.$

To find the gradient, we need to:

1. Differentiate $y$ with respect to $x$.
2. Evaluate the derivative at $x = 1$.

Step 1: Differentiate $y = \frac{\sqrt{2x - 1}}{x}$

Using the quotient rule for differentiation:

$\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2},$

where:

• $u = \sqrt{2x - 1}$,
• $v = x$.

Differentiate $u$ and $v$:

• $\frac{du}{dx} = \frac{d}{dx} \sqrt{2x - 1} = \frac{1}{2\sqrt{2x - 1}} \cdot 2 = \frac{1}{\sqrt{2x - 1}}$,
• $\frac{dv}{dx} = 1$.

Substitute into the quotient rule:

$\frac{dy}{dx} = \frac{x \cdot \frac{1}{\sqrt{2x - 1}} - \sqrt{2x - 1} \cdot 1}{x^2}.$

Simplify: $\frac{dy}{dx} = \frac{\frac{x}{\sqrt{2x - 1}} - \sqrt{2x - 1}}{x^2}.$

Combine the terms in the numerator under a common denominator: $\frac{dy}{dx} = \frac{\frac{x - (2x - 1)}{\sqrt{2x - 1}}}{x^2}.$

Simplify further: $\frac{dy}{dx} = \frac{\frac{-x + 1}{\sqrt{2x - 1}}}{x^2}.$

This becomes: $\frac{dy}{dx} = \frac{1 - x}{x^2 \sqrt{2x - 1}}.$

Step 2: Evaluate the derivative at $x = 1$

Substitute $x = 1$ into the derivative: $\frac{dy}{dx} = \frac{1 - 1}{1^2 \sqrt{2(1) - 1}} = \frac{0}{1 \cdot \sqrt{1}} = 0.$

Final Answer:

The gradient of the tangent to the curve at $x = 1$ is:

$\boxed{0}.$

Let me know if you'd like more details or have any other questions!

Related Questions:

1. How is the quotient rule derived?
2. Why does the numerator $1 - x$ simplify to zero at $x = 1$?
3. What happens to the curve when $x < \frac{1}{2}$, and why is $\sqrt{2x - 1}$ undefined there?
4. How can you confirm the gradient calculation graphically?
5. What is the physical meaning of a gradient of zero in this context?

Tip: Always check the domain of the function before differentiating; in this case, $2x - 1 \geq 0$, so $x \geq \frac{1}{2}$.