If : `f(x)=(sin^(2)x)/(1+cotx)+(cos^(2)x)/(1+tanx)," then: "f'((pi)/(4))=`

Options

A. `sqrt(2)`
B. `1//sqrt(3)`
C. 0
D. `-sqrt(2)`

Answer

To find \( f'\left(\frac{\pi}{4}\right) \) for the function

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

we will follow these steps:

### Step 1: Rewrite the function
We start by rewriting the cotangent and tangent functions in terms of sine and cosine:

\[ \cot x = \frac{\cos x}{\sin x} \quad \text{and} \quad \tan x = \frac{\sin x}{\cos x}. \]

Thus, we can rewrite \( f(x) \):

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

### Step 2: Simplify the fractions
Now, we simplify each term:

1. For the first term:
\[ \frac{\sin^2 x}{1 + \frac{\cos x}{\sin x}} = \frac{\sin^2 x}{\frac{\sin x + \cos x}{\sin x}} = \frac{\sin^3 x}{\sin x + \cos x}. \]

2. For the second term:
\[ \frac{\cos^2 x}{1 + \frac{\sin x}{\cos x}} = \frac{\cos^2 x}{\frac{\cos x + \sin x}{\cos x}} = \frac{\cos^3 x}{\sin x + \cos x}. \]

Combining both terms, we have:

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

### Step 3: Use the identity for sum of cubes
Recall the identity for the sum of cubes:

\[ a^3 + b^3 = (a + b)(a^2 - ab + b^2). \]

Applying this to our function:

\[ \sin^3 x + \cos^3 x = (\sin x + \cos x)(\sin^2 x - \sin x \cos x + \cos^2 x). \]

Thus, we can simplify \( f(x) \):

\[ f(x) = \sin^2 x - \sin x \cos x + \cos^2 x. \]

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

\[ f(x) = 1 - \sin x \cos x. \]

### Step 4: Differentiate \( f(x) \)
Now we differentiate \( f(x) \):

\[ f'(x) = 0 - \frac{d}{dx}(\sin x \cos x). \]

Using the product rule:

\[ \frac{d}{dx}(\sin x \cos x) = \sin x \frac{d}{dx}(\cos x) + \cos x \frac{d}{dx}(\sin x) = \sin x (-\sin x) + \cos x (\cos x) = \cos^2 x - \sin^2 x. \]

Thus,

\[ f'(x) = -(\cos^2 x - \sin^2 x) = \sin^2 x - \cos^2 x. \]

### Step 5: Evaluate \( f'\left(\frac{\pi}{4}\right) \)
Now we evaluate \( f'\left(\frac{\pi}{4}\right) \):

At \( x = \frac{\pi}{4} \):

\[ \sin\left(\frac{\pi}{4}\right) = \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}. \]

Thus,

\[ f'\left(\frac{\pi}{4}\right) = \left(\frac{1}{\sqrt{2}}\right)^2 - \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2} - \frac{1}{2} = 0. \]

### Final Answer
\[ f'\left(\frac{\pi}{4}\right) = 0. \]