If `y = log_(cos x) sin x " then" (dy)/(dx)` is equal to

Options

A. `((cot x log cos x + tan x log sin x))/((log cos x)^(2))`
B. `((tan x log cos x + cot x log sin x))/((log cos x)^(2))`
C. `((cot x log cos x + tan x log sin x))/((log sin x)^(2))`
D. None of the above

Explanation

To find the derivative of the function \( y = \log_{\cos x} (\sin x) \), we can use the change of base formula for logarithms. The change of base formula states that:

\[ \log_a b = \frac{\log b}{\log a} \]

Using this, we can rewrite \( y \) as:

\[ y = \frac{\log(\sin x)}{\log(\cos x)} \]

Now, we will differentiate \( y \) using the quotient rule. The quotient rule states that if you have a function \( \frac{u}{v} \), its derivative is given by:

\[ \frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]

Here, let:

- \( u = \log(\sin x) \)
- \( v = \log(\cos x) \)

Now, we need to find \( \frac{du}{dx} \) and \( \frac{dv}{dx} \).

1. **Finding \( \frac{du}{dx} \)**:

Using the chain rule, we have:

\[ \frac{du}{dx} = \frac{1}{\sin x} \cdot \cos x = \cot x \]

2. **Finding \( \frac{dv}{dx} \)**:

Similarly, for \( v \):

\[ \frac{dv}{dx} = \frac{1}{\cos x} \cdot (-\sin x) = -\tan x \]

Now substituting \( u \), \( v \), \( \frac{du}{dx} \), and \( \frac{dv}{dx} \) into the quotient rule formula:

\[ \frac{dy}{dx} = \frac{\log(\cos x) \cdot \cot x - \log(\sin x) \cdot (-\tan x)}{(\log(\cos x))^2} \]

This simplifies to:

\[ \frac{dy}{dx} = \frac{\log(\cos x) \cdot \cot x + \log(\sin x) \cdot \tan x}{(\log(\cos x))^2} \]

Thus, the final answer is:

\[ \frac{dy}{dx} = \frac{\log(\cos x) \cdot \cot x + \log(\sin x) \cdot \tan x}{(\log(\cos x))^2} \]