`(d)/(dx)[log(sqrt((1+sinx)/(1-sinx)))]=`

Options

A. `cscx`
B. `tanx`
C. `cosx`
D. `secx`

Answer

To solve the problem \(\frac{d}{dx}\left[\log\left(\sqrt{\frac{1+\sin x}{1-\sin x}}\right)\right]\), we will follow these steps:

### Step 1: Simplify the expression inside the logarithm
We start with the expression:
\[ \log\left(\sqrt{\frac{1+\sin x}{1-\sin x}}\right) \]
Using the property of logarithms, \(\log(\sqrt{a}) = \frac{1}{2}\log(a)\), we can rewrite it as:
\[ \frac{1}{2} \log\left(\frac{1+\sin x}{1-\sin x}\right) \]

### Step 2: Differentiate the logarithmic expression
Now, we differentiate:
\[ \frac{d}{dx}\left[\frac{1}{2} \log\left(\frac{1+\sin x}{1-\sin x}\right)\right] \]
Using the chain rule, we have:
\[ \frac{1}{2} \cdot \frac{1}{\frac{1+\sin x}{1-\sin x}} \cdot \frac{d}{dx}\left(\frac{1+\sin x}{1-\sin x}\right) \]

### Step 3: Differentiate the fraction \(\frac{1+\sin x}{1-\sin x}\)
We will use the quotient rule for differentiation, which states that if \(u = 1 + \sin x\) and \(v = 1 - \sin x\), then:
\[ \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2} \]
Calculating \(u'\) and \(v'\):
- \(u' = \cos x\)
- \(v' = -\cos x\)

Now applying the quotient rule:
\[ \frac{d}{dx}\left(\frac{1+\sin x}{1-\sin x}\right) = \frac{(\cos x)(1 - \sin x) - (1 + \sin x)(-\cos x)}{(1 - \sin x)^2} \]
Simplifying the numerator:
\[ \cos x(1 - \sin x) + \cos x(1 + \sin x) = \cos x(1 - \sin x + 1 + \sin x) = \cos x(2) = 2\cos x \]
Thus, we have:
\[ \frac{d}{dx}\left(\frac{1+\sin x}{1-\sin x}\right) = \frac{2\cos x}{(1 - \sin x)^2} \]

### Step 4: Substitute back into the derivative
Now substituting back into our derivative:
\[ \frac{1}{2} \cdot \frac{1}{\frac{1+\sin x}{1-\sin x}} \cdot \frac{2\cos x}{(1 - \sin x)^2} \]
This simplifies to:
\[ \frac{1}{2} \cdot \frac{1 - \sin x}{1 + \sin x} \cdot \frac{2\cos x}{(1 - \sin x)^2} = \frac{\cos x (1 - \sin x)}{(1 + \sin x)(1 - \sin x)^2} \]
The \((1 - \sin x)\) cancels out:
\[ \frac{\cos x}{(1 + \sin x)(1 - \sin x)} = \frac{\cos x}{\cos^2 x} = \sec x \]

### Final Answer
Thus, the final answer is:
\[ \frac{d}{dx}\left[\log\left(\sqrt{\frac{1+\sin x}{1-\sin x}}\right)\right] = \sec x \]