Differentiate the following w.r.t.x.
`(cosx)^(x)`

Answer

To differentiate the function \( y = (\cos x)^x \) with respect to \( x \), we will use logarithmic differentiation. Here’s the step-by-step solution:

### Step 1: Take the natural logarithm of both sides
We start by taking the natural logarithm of both sides:
\[ \ln y = \ln((\cos x)^x) \]

### Step 2: Apply the power rule of logarithms
Using the property of logarithms that states \( \ln(a^b) = b \ln a \), we rewrite the equation:
\[ \ln y = x \ln(\cos x) \]

### Step 3: Differentiate both sides
Now we differentiate both sides with respect to \( x \). We will use implicit differentiation on the left side and the product rule on the right side:
\[ \frac{d}{dx}(\ln y) = \frac{d}{dx}(x \ln(\cos x)) \]

Using the chain rule on the left side:
\[ \frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}(x \ln(\cos x)) \]

### Step 4: Differentiate the right side using the product rule
Now we apply the product rule on the right side:
\[ \frac{d}{dx}(x \ln(\cos x)) = \ln(\cos x) \cdot \frac{d}{dx}(x) + x \cdot \frac{d}{dx}(\ln(\cos x)) \]
\[ = \ln(\cos x) + x \cdot \frac{1}{\cos x} \cdot \frac{d}{dx}(\cos x) \]

### Step 5: Differentiate \(\cos x\)
The derivative of \(\cos x\) is \(-\sin x\):
\[ = \ln(\cos x) + x \cdot \frac{1}{\cos x} \cdot (-\sin x) \]
\[ = \ln(\cos x) - x \frac{\sin x}{\cos x} \]
\[ = \ln(\cos x) - x \tan x \]

### Step 6: Substitute back into the equation
Now we substitute this back into our differentiated equation:
\[ \frac{1}{y} \frac{dy}{dx} = \ln(\cos x) - x \tan x \]

### Step 7: Solve for \(\frac{dy}{dx}\)
To find \(\frac{dy}{dx}\), we multiply both sides by \( y \):
\[ \frac{dy}{dx} = y \left(\ln(\cos x) - x \tan x\right) \]

### Step 8: Substitute back for \( y \)
Since \( y = (\cos x)^x \), we substitute back:
\[ \frac{dy}{dx} = (\cos x)^x \left(\ln(\cos x) - x \tan x\right) \]

### Final Answer
Thus, the derivative of \( (\cos x)^x \) with respect to \( x \) is:
\[ \frac{dy}{dx} = (\cos x)^x \left(\ln(\cos x) - x \tan x\right) \]

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