If `f(x)=(1-sin^(2)x)/(1+sin^(2)x)," then "3f'((pi)/(4))-f((pi)/(4))=`

Options

A. `-3`
B. 0
C. 1
D. 3

Answer

To solve the problem, we need to find \( 3f'(\frac{\pi}{4}) - f(\frac{\pi}{4}) \) where \( f(x) = \frac{1 - \sin^2 x}{1 + \sin^2 x} \).

### Step 1: Simplify the function \( f(x) \)

We can simplify \( f(x) \) using the identity \( \sin^2 x + \cos^2 x = 1 \):

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

### Step 2: Differentiate \( f(x) \)

To differentiate \( f(x) \), we will use the quotient rule, which states:

\[ \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \]

Here, let \( u = \cos^2 x \) and \( v = 1 + \sin^2 x \).

1. Differentiate \( u \):
\[ u' = 2\cos x(-\sin x) = -2\sin x \cos x \]

2. Differentiate \( v \):
\[ v' = 2\sin x \cos x \]

Now applying the quotient rule:

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

### Step 3: Simplify \( f'(x) \)

Combining the terms in the numerator:

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

Factoring out \( -2\sin x \cos x \):

\[ = \frac{-2\sin x \cos x \left( (1 + \sin^2 x) + \cos^2 x \right)}{(1 + \sin^2 x)^2} \]

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

\[ = \frac{-2\sin x \cos x \cdot 2}{(1 + \sin^2 x)^2} = \frac{-4\sin x \cos x}{(1 + \sin^2 x)^2} \]

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

Now we evaluate \( f'(\frac{\pi}{4}) \):

1. Calculate \( \sin(\frac{\pi}{4}) = \cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}} \).

2. Substitute into \( f'(\frac{\pi}{4}) \):

\[ f'(\frac{\pi}{4}) = \frac{-4 \cdot \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}}}{(1 + \frac{1}{2})^2} = \frac{-4 \cdot \frac{1}{2}}{\left(\frac{3}{2}\right)^2} = \frac{-2}{\frac{9}{4}} = \frac{-8}{9} \]

### Step 5: Evaluate \( f(\frac{\pi}{4}) \)

Now we evaluate \( f(\frac{\pi}{4}) \):

\[ f(\frac{\pi}{4}) = \frac{\cos^2(\frac{\pi}{4})}{1 + \sin^2(\frac{\pi}{4})} = \frac{\left(\frac{1}{\sqrt{2}}\right)^2}{1 + \left(\frac{1}{\sqrt{2}}\right)^2} = \frac{\frac{1}{2}}{1 + \frac{1}{2}} = \frac{\frac{1}{2}}{\frac{3}{2}} = \frac{1}{3} \]

### Step 6: Calculate \( 3f'(\frac{\pi}{4}) - f(\frac{\pi}{4}) \)

Finally, we compute:

\[ 3f'(\frac{\pi}{4}) - f(\frac{\pi}{4}) = 3 \left(\frac{-8}{9}\right) - \frac{1}{3} \]

Calculating \( 3 \cdot \frac{-8}{9} = \frac{-24}{9} \) and converting \( \frac{1}{3} \) to a fraction with a denominator of 9:

\[ \frac{1}{3} = \frac{3}{9} \]

Thus,

\[ 3f'(\frac{\pi}{4}) - f(\frac{\pi}{4}) = \frac{-24}{9} - \frac{3}{9} = \frac{-27}{9} = -3 \]

### Final Answer

\[ 3f'(\frac{\pi}{4}) - f(\frac{\pi}{4}) = -3 \]