If `y=(cosx)^(cosx)^(cosx)^^((((oo)))),p rov et h a t(dy)/(dx)=-(y^2tanx)/((1-ylogcosx)dot`

Explanation

To solve the problem, we start with the expression given for \( y \):

\[ y = (\cos x)^{(\cos x)^{(\cos x)^{\cdots}}} \]

This expression can be rewritten using the property of infinite exponentials. We can assume that the infinite exponent is equal to \( y \):

\[ y = \cos x^y \]

### Step 1: Take the logarithm of both sides

Taking the logarithm of both sides gives:

\[ \log y = \log(\cos x^y) \]

Using the logarithmic identity \( \log(a^b) = b \log a \), we can rewrite the right side:

\[ \log y = y \log(\cos x) \]

### Step 2: Differentiate both sides with respect to \( x \)

Now we differentiate both sides with respect to \( x \):

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

Using the chain rule on the left side:

\[ \frac{1}{y} \frac{dy}{dx} \]

For the right side, we apply the product rule:

\[ \frac{d}{dx}(y \log(\cos x)) = \frac{dy}{dx} \log(\cos x) + y \frac{d}{dx}(\log(\cos x)) \]

The derivative of \( \log(\cos x) \) is:

\[ \frac{d}{dx}(\log(\cos x)) = -\tan x \]

Thus, we have:

\[ \frac{1}{y} \frac{dy}{dx} = \frac{dy}{dx} \log(\cos x) - y \tan x \]

### Step 3: Rearranging the equation

Rearranging the equation gives:

\[ \frac{1}{y} \frac{dy}{dx} - \frac{dy}{dx} \log(\cos x) = -y \tan x \]

Factoring out \( \frac{dy}{dx} \) from the left side:

\[ \frac{dy}{dx} \left( \frac{1}{y} - \log(\cos x) \right) = -y \tan x \]

### Step 4: Solve for \( \frac{dy}{dx} \)

Now, we can solve for \( \frac{dy}{dx} \):

\[ \frac{dy}{dx} = -y \tan x \cdot \frac{1}{\frac{1}{y} - \log(\cos x)} \]

This simplifies to:

\[ \frac{dy}{dx} = -y^2 \tan x \cdot \frac{1}{1 - y \log(\cos x)} \]

### Final Result

Thus, we have proven that:

\[ \frac{dy}{dx} = -\frac{y^2 \tan x}{1 - y \log(\cos x)} \]