The function f whose graph passes through `(pi//4, 0)` and whose derivative is `log(tanx)/(sinxcosx)` is given by

Options

A. `(log tanx)^(2)+(pi)/(4)`
B. `(1)/(2)(logcosx)^(2)`
C. `(1)/(2)(logsinx)^(2)`
D. `(1)/(2)(logtanx)^(2)`

Answer

To find the function \( f(x) \) whose derivative is given by

\[ f'(x) = \frac{\log(\tan x)}{\sin x \cos x} \]

and which passes through the point \( \left(\frac{\pi}{4}, 0\right) \), we will follow these steps:

### Step 1: Rewrite the derivative
We start with the given derivative:

\[ f'(x) = \frac{\log(\tan x)}{\sin x \cos x} \]

We can express this in differential form:

\[ df = \frac{\log(\tan x)}{\sin x \cos x} \, dx \]

### Step 2: Integrate both sides
Next, we integrate both sides to find \( f(x) \):

\[ f(x) = \int \frac{\log(\tan x)}{\sin x \cos x} \, dx \]

### Step 3: Use substitution
To make the integration easier, we can use the substitution:

Let \( t = \log(\tan x) \). Then, we differentiate \( t \):

\[ dt = \frac{1}{\tan x} \cdot \sec^2 x \, dx = \frac{1}{\sin x \cos x} \, dx \]

Thus, we can express \( dx \) in terms of \( dt \):

\[ dx = \sin x \cos x \, dt \]

### Step 4: Substitute and simplify
Substituting \( t \) into the integral gives us:

\[ f(x) = \int t \, dt \]

### Step 5: Integrate
The integral of \( t \) is:

\[ f(x) = \frac{t^2}{2} + C = \frac{(\log(\tan x))^2}{2} + C \]

### Step 6: Apply the initial condition
We know that the function passes through the point \( \left(\frac{\pi}{4}, 0\right) \). Thus, we substitute \( x = \frac{\pi}{4} \) and \( f\left(\frac{\pi}{4}\right) = 0 \):

\[ 0 = \frac{(\log(\tan(\frac{\pi}{4})))^2}{2} + C \]

Since \( \tan\left(\frac{\pi}{4}\right) = 1 \), we have \( \log(1) = 0 \):

\[ 0 = \frac{0^2}{2} + C \implies C = 0 \]

### Step 7: Final function
Thus, the function \( f(x) \) is:

\[ f(x) = \frac{(\log(\tan x))^2}{2} \]

### Final Answer
The function \( f(x) \) is given by:

\[ f(x) = \frac{1}{2} \log^2(\tan x) \]