(b) (i) Differentiate \( y=\tan (x-\sqrt{x}) \) with respect to \( x \). (ii) Hence find \( \int \frac{2 \sqrt{x}-1}{\sqrt{x} \cos ^{2}(x-\sqrt{x})} \mathrm{d} x \)

To solve the problem, we will first differentiate the function \( y = \tan(x - \sqrt{x}) \) with respect to \( x \), and then we will use that result to evaluate the integral. ### (b) (i) Differentiate \( y = \tan(x - \sqrt{x}) \) Using the chain rule, we have: \[ \frac{dy}{dx} = \sec^2(x - \sqrt{x}) \cdot \frac{d}{dx}(x - \sqrt{x}) \] Now, we need to differentiate \( x - \sqrt{x} \): \[ \frac{d}{dx}(x - \sqrt{x}) = 1 - \frac{1}{2\sqrt{x}} \] Thus, we can substitute this back into our derivative: \[ \frac{dy}{dx} = \sec^2(x - \sqrt{x}) \left( 1 - \frac{1}{2\sqrt{x}} \right) \] ### (b) (ii) Find \( \int \frac{2 \sqrt{x} - 1}{\sqrt{x} \cos^2(x - \sqrt{x})} \, dx \) We can use the result from part (i). Notice that: \[ \frac{dy}{dx} = \sec^2(x - \sqrt{x}) \left( 1 - \frac{1}{2\sqrt{x}} \right) \] Rearranging gives us: \[ \sec^2(x - \sqrt{x}) = \frac{dy/dx}{1 - \frac{1}{2\sqrt{x}}} \] Now, we can express \( \cos^2(x - \sqrt{x}) \) in terms of \( \sec^2(x - \sqrt{x}) \): \[ \cos^2(x - \sqrt{x}) = \frac{1}{\sec^2(x - \sqrt{x})} \] Thus, we can rewrite the integral: \[ \int \frac{2 \sqrt{x} - 1}{\sqrt{x} \cos^2(x - \sqrt{x})} \, dx = \int (2\sqrt{x} - 1) \cdot \frac{\sec^2(x - \sqrt{x})}{\sqrt{x}} \, dx \] Now, we can substitute \( \sec^2(x - \sqrt{x}) \): \[ = \int (2\sqrt{x} - 1) \cdot \frac{dy/dx}{\sqrt{x} \left( 1 - \frac{1}{2\sqrt{x}} \right)} \, dx \] This integral can be simplified further. Notice that: \[ \frac{2\sqrt{x} - 1}{\sqrt{x}} = 2 - \frac{1}{\sqrt{x}} \] Thus, we can split the integral: \[ = \int \left( 2 - \frac{1}{\sqrt{x}} \right) \cdot \frac{dy/dx}{1 - \frac{1}{2\sqrt{x}}} \, dx \] This integral can be evaluated by recognizing that the expression \( \frac{dy}{dx} \) is the derivative of \( y \). Therefore, we can integrate directly: \[ \int \frac{dy}{dx} \, dx = y + C \] Thus, we have: \[ \int \frac{2 \sqrt{x} - 1}{\sqrt{x} \cos^2(x - \sqrt{x})} \, dx = \tan(x - \sqrt{x}) + C \] ### Final Answer \[ \int \frac{2 \sqrt{x} - 1}{\sqrt{x} \cos^2(x - \sqrt{x})} \, dx = \tan(x - \sqrt{x}) + C \]