Differentiate w.r.t `x`
(i) `(3x^(2) +4x-5)/x`
(ii) ` ((x^(3) +1)(x-2))/x^(2)`
(iii) ` (x-4)/(2sqrtx) `
(iv) ` (( 1+x)sqrtx)/root3(x)`
(v) `(ax^(2) +bx+c)/sqrtx `
(vi) ` (a+b cos x)/sin x `

Answer

Let's solve each part of the question step by step.

### (i) Differentiate \( \frac{3x^2 + 4x - 5}{x} \)

1. **Rewrite the expression**:
\[ \frac{3x^2 + 4x - 5}{x} = 3x + 4 - \frac{5}{x} \]

2. **Differentiate**:
\[ \frac{dy}{dx} = \frac{d}{dx}(3x) + \frac{d}{dx}(4) - \frac{d}{dx}\left(\frac{5}{x}\right) \]
\[ = 3 + 0 - \frac{d}{dx}(5x^{-1}) \]
\[ = 3 + 0 + 5x^{-2} \]
\[ = 3 + \frac{5}{x^2} \]

**Final Answer**:
\[ \frac{dy}{dx} = 3 + \frac{5}{x^2} \]

### (ii) Differentiate \( \frac{(x^3 + 1)(x - 2)}{x^2} \)

1. **Rewrite the expression**:
\[ y = \frac{(x^3 + 1)(x - 2)}{x^2} \]

2. **Use the product rule**:
Let \( u = (x^3 + 1) \) and \( v = (x - 2) \).
\[ y = \frac{u \cdot v}{x^2} \]

3. **Differentiate using the quotient rule**:
\[ \frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{x^2} \cdot \frac{1}{x^2} \]
where \( \frac{du}{dx} = 3x^2 \) and \( \frac{dv}{dx} = 1 \).

4. **Substitute and simplify**:
\[ = \frac{(x - 2)(3x^2) - (x^3 + 1)(1)}{x^4} \]
\[ = \frac{3x^3 - 6x^2 - x^3 - 1}{x^4} \]
\[ = \frac{2x^3 - 6x^2 - 1}{x^4} \]

**Final Answer**:
\[ \frac{dy}{dx} = \frac{2x^3 - 6x^2 - 1}{x^4} \]

### (iii) Differentiate \( \frac{x - 4}{2\sqrt{x}} \)

1. **Rewrite the expression**:
\[ y = \frac{x - 4}{2x^{1/2}} = \frac{1}{2}(x^{1/2} - 4x^{-1/2}) \]

2. **Differentiate**:
\[ \frac{dy}{dx} = \frac{1}{2}\left(\frac{1}{2}x^{-1/2} + 2 \cdot 4x^{-3/2}\right) \]
\[ = \frac{1}{4\sqrt{x}} + \frac{8}{2x^{3/2}} = \frac{1}{4\sqrt{x}} + \frac{4}{x^{3/2}} \]

**Final Answer**:
\[ \frac{dy}{dx} = \frac{1}{4\sqrt{x}} + \frac{4}{x^{3/2}} \]

### (iv) Differentiate \( \frac{(1 + x)\sqrt{x}}{\sqrt[3]{x}} \)

1. **Rewrite the expression**:
\[ y = (1 + x)x^{1/2}x^{-1/3} = (1 + x)x^{1/6} \]

2. **Use the product rule**:
Let \( u = (1 + x) \) and \( v = x^{1/6} \).
\[ \frac{dy}{dx} = u'v + uv' \]
where \( u' = 1 \) and \( v' = \frac{1}{6}x^{-5/6} \).

3. **Substitute and simplify**:
\[ = (1)(x^{1/6}) + (1 + x)\left(\frac{1}{6}x^{-5/6}\right) \]
\[ = x^{1/6} + \frac{1 + x}{6x^{5/6}} \]

**Final Answer**:
\[ \frac{dy}{dx} = x^{1/6} + \frac{1 + x}{6x^{5/6}} \]

### (v) Differentiate \( \frac{ax^2 + bx + c}{\sqrt{x}} \)

1. **Rewrite the expression**:
\[ y = (ax^2 + bx + c)x^{-1/2} \]

2. **Use the product rule**:
Let \( u = ax^2 + bx + c \) and \( v = x^{-1/2} \).
\[ \frac{dy}{dx} = u'v + uv' \]
where \( u' = 2ax + b \) and \( v' = -\frac{1}{2}x^{-3/2} \).

3. **Substitute and simplify**:
\[ = (2ax + b)x^{-1/2} + (ax^2 + bx + c)\left(-\frac{1}{2}x^{-3/2}\right) \]

**Final Answer**:
\[ \frac{dy}{dx} = (2ax + b)x^{-1/2} - \frac{1}{2}(ax^2 + bx + c)x^{-3/2} \]

### (vi) Differentiate \( \frac{a + b \cos x}{\sin x} \)

1. **Use the quotient rule**:
Let \( u = a + b \cos x \) and \( v = \sin x \).
\[ \frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]
where \( \frac{du}{dx} = -b \sin x \) and \( \frac{dv}{dx} = \cos x \).

2. **Substitute and simplify**:
\[ = \frac{\sin x(-b \sin x) - (a + b \cos x)(\cos x)}{\sin^2 x} \]
\[ = \frac{-b \sin^2 x - a \cos x - b \cos^2 x}{\sin^2 x} \]

**Final Answer**:
\[ \frac{dy}{dx} = \frac{-b \sin^2 x - a \cos x - b \cos^2 x}{\sin^2 x} \]

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