If `f(x)=(2-3cosx)/(sinx)`, then `f'((pi)/(4))` is equal to

Options

A. `2sqrt(2)-6`
B. `6-2sqrt(2)`
C. `3-sqrt(2)`
D. `sqrt(2)-3`

Answer

To find \( f'\left(\frac{\pi}{4}\right) \) for the function \( f(x) = \frac{2 - 3\cos x}{\sin x} \), we will follow these steps:

### Step 1: Find the derivative \( f'(x) \)

Using the quotient rule, which states that if \( f(x) = \frac{g(x)}{h(x)} \), then

\[ f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{[h(x)]^2} \]

In our case, \( g(x) = 2 - 3\cos x \) and \( h(x) = \sin x \).

#### Step 1.1: Calculate \( g'(x) \) and \( h'(x) \)

- \( g'(x) = 0 + 3\sin x = 3\sin x \) (since the derivative of \( -3\cos x \) is \( 3\sin x \))
- \( h'(x) = \cos x \)

#### Step 1.2: Apply the quotient rule

Now substituting into the quotient rule:

\[ f'(x) = \frac{(3\sin x)(\sin x) - (2 - 3\cos x)(\cos x)}{(\sin x)^2} \]

### Step 2: Simplify the expression

Expanding the numerator:

\[ f'(x) = \frac{3\sin^2 x - (2\cos x - 3\cos^2 x)}{\sin^2 x} \]

This simplifies to:

\[ f'(x) = \frac{3\sin^2 x + 3\cos^2 x - 2\cos x}{\sin^2 x} \]

Using the identity \( \sin^2 x + \cos^2 x = 1 \):

\[ f'(x) = \frac{3(1) - 2\cos x}{\sin^2 x} = \frac{3 - 2\cos x}{\sin^2 x} \]

### Step 3: Evaluate \( f'\left(\frac{\pi}{4}\right) \)

Now we substitute \( x = \frac{\pi}{4} \):

\[ f'\left(\frac{\pi}{4}\right) = \frac{3 - 2\cos\left(\frac{\pi}{4}\right)}{\sin^2\left(\frac{\pi}{4}\right)} \]

Knowing that \( \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \) and \( \sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \):

\[ f'\left(\frac{\pi}{4}\right) = \frac{3 - 2 \cdot \frac{1}{\sqrt{2}}}{\left(\frac{1}{\sqrt{2}}\right)^2} \]

Calculating \( \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2} \):

\[ f'\left(\frac{\pi}{4}\right) = \frac{3 - \frac{2}{\sqrt{2}}}{\frac{1}{2}} = 2(3 - \frac{2}{\sqrt{2}}) \]

### Step 4: Simplify further

\[ = 6 - 2\sqrt{2} \]

Thus, the final answer is:

\[ f'\left(\frac{\pi}{4}\right) = 6 - 2\sqrt{2} \]