Differentiate the following functions: h) u = (2/√x - 1)^(-1), i) v = (2x2 − 3/x2)^(1/3)

let's differentiate these functions.

h) $u = \left(\frac{2}{\sqrt{x}} - 1\right)^{-1}$

Step 1: Rewrite the function using fractional and negative exponents.

$$u = (2x^{-1/2} - 1)^{-1}$$

Step 2: Apply the chain rule. Let $w = 2x^{-1/2} - 1$.

$$\frac{du}{dx} = -1(2x^{-1/2} - 1)^{-1-1} \cdot \frac{d}{dx}(2x^{-1/2} - 1)$$

Step 3: Differentiate the inner function.

$$\frac{d}{dx}(2x^{-1/2} - 1) = 2 \left(-\frac{1}{2}x^{-\frac{1}{2}-1}\right) - 0 = -x^{-3/2}$$

Step 4: Substitute and simplify.

$$\frac{du}{dx} = -1(2x^{-1/2} - 1)^{-2} (-x^{-3/2}) = x^{-3/2}(2x^{-1/2} - 1)^{-2}$$

Step 5: Rewrite with positive and radical exponents.

$$\frac{du}{dx} = \frac{1}{x^{3/2}\left(\frac{2}{\sqrt{x}} - 1\right)^2} = \frac{1}{x\sqrt{x}\left(\frac{2-\sqrt{x}}{\sqrt{x}}\right)^2} = \frac{1}{x\sqrt{x}\frac{(2-\sqrt{x})^2}{x}} = \frac{1}{\sqrt{x}(2-\sqrt{x})^2}$$

The derivative is $\boxed{\mathbf{\frac{1}{\sqrt{x}(2-\sqrt{x})^2}}}$.

i) $v = \left(2x^2 - \frac{3}{x^2}\right)^{1/3}$

Step 1: Rewrite the function using negative exponents.

$$v = (2x^2 - 3x^{-2})^{1/3}$$

Step 2: Apply the chain rule. Let $w = 2x^2 - 3x^{-2}$.

$$\frac{dv}{dx} = \frac{1}{3}(2x^2 - 3x^{-2})^{\frac{1}{3}-1} \cdot \frac{d}{dx}(2x^2 - 3x^{-2})$$

Step 3: Differentiate the inner function.

$$\frac{d}{dx}(2x^2 - 3x^{-2}) = 2(2x) - 3(-2x^{-2-1}) = 4x + 6x^{-3}$$

Step 4: Substitute and simplify.

$$\frac{dv}{dx} = \frac{1}{3}(2x^2 - 3x^{-2})^{-2/3} (4x + 6x^{-3})$$

Step 5: Rewrite with positive exponents.

$$\frac{dv}{dx} = \frac{4x + \frac{6}{x^3}}{3\left(2x^2 - \frac{3}{x^2}\right)^{2/3}} = \frac{\frac{4x^4+6}{x^3}}{3\left(\frac{2x^4-3}{x^2}\right)^{2/3}} = \frac{2(2x^4+3)}{3x^3\left(2x^2 - \frac{3}{x^2}\right)^{2/3}}$$

The derivative is $\boxed{\mathbf{\frac{2(2x^4+3)}{3x^3\left(2x^2 - \frac{3}{x^2}\right)^{2/3}}}}$.

j) $y = \sqrt{x - \frac{1}{x}}$

Step 1: Rewrite the function using fractional and negative exponents.

$$y = (x - x^{-1})^{1/2}$$

Step 2: Apply the chain rule. Let $w = x - x^{-1}$.

$$\frac{dy}{dx} = \frac{1}{2}(x - x^{-1})^{\frac{1}{2}-1} \cdot \frac{d}{dx}(x - x^{-1})$$

Step 3: Differentiate the inner function.

$$\frac{d}{dx}(x - x^{-1}) = 1 - (-1x^{-1-1}) = 1 + x^{-2}$$

Step 4: Substitute and simplify.

$$\frac{dy}{dx} = \frac{1}{2}(x - x^{-1})^{-1/2} (1 + x^{-2})$$

Step 5: Rewrite with positive and radical exponents.

$$\frac{dy}{dx} = \frac{1 + \frac{1}{x^2}}{2\sqrt{x - \frac{1}{x}}} = \frac{\frac{x^2+1}{x^2}}{2\sqrt{\frac{x^2-1}{x}}} = \frac{x^2+1}{2x^2 \frac{\sqrt{x^2-1}}{\sqrt{x}}} = \frac{x^2+1}{2x^{3/2}\sqrt{x^2-1}}$$

The derivative is $\boxed{\mathbf{\frac{x^2+1}{2x^{3/2}\sqrt{x^2-1}}}}$.

k) $y = \left(\sqrt{x} - \frac{1}{\sqrt{x}}\right)^2$

Step 1: Rewrite the function using fractional exponents.

$$y = (x^{1/2} - x^{-1/2})^2$$

Step 2: Expand the expression first, as it simplifies nicely.

$$y = (x^{1/2})^2 - 2(x^{1/2})(x^{-1/2}) + (x^{-1/2})^2$$ $$y = x - 2x^0 + x^{-1}$$ $$y = x - 2 + x^{-1}$$

Step 3: Differentiate each term using the power rule.

$$\frac{dy}{dx} = 1 - 0 - 1x^{-1-1} = 1 - x^{-2}$$

Step 4: Rewrite with positive exponents.

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

The derivative is $\boxed{\mathbf{1 - \frac{1}{x^2}}}$.

l) $y = \sqrt[3]{(1 - \sqrt{x})}$

Step 1: Rewrite the function using fractional exponents.

$$y = (1 - x^{1/2})^{1/3}$$

Step 2: Apply the chain rule. Let $w = 1 - x^{1/2}$.

$$\frac{dy}{dx} = \frac{1}{3}(1 - x^{1/2})^{\frac{1}{3}-1} \cdot \frac{d}{dx}(1 - x^{1/2})$$

Step 3: Differentiate the inner function.

$$\frac{d}{dx}(1 - x^{1/2}) = 0 - \frac{1}{2}x^{\frac{1}{2}-1} = -\frac{1}{2}x^{-1/2}$$

Step 4: Substitute and simplify.

$$\frac{dy}{dx} = \frac{1}{3}(1 - x^{1/2})^{-2/3} \left(-\frac{1}{2}x^{-1/2}\right) = -\frac{1}{6}x^{-1/2}(1 - x^{1/2})^{-2/3}$$

Step 5: Rewrite with positive and radical exponents.

$$\frac{dy}{dx} = -\frac{1}{6\sqrt{x}(1 - \sqrt{x})^{2/3}}$$

The derivative is $\boxed{\mathbf{-\frac{1}{6\sqrt{x}(1 - \sqrt{x})^{2/3}}}}$.

m) $y = \frac{1}{x^{2/3} - 1}$

Step 1: Rewrite the function using negative exponents.

$$y = (x^{2/3} - 1)^{-1}$$

Step 2: Apply the chain rule. Let $w = x^{2/3} - 1$.

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