Differentiate the following w.r.t. x :
`1/(logcosx)`

Answer

To differentiate the function \( y = \frac{1}{\log(\cos x)} \) with respect to \( x \), we will use the quotient rule of differentiation. The quotient rule states that if you have a function in the form \( \frac{u}{v} \), the derivative is given by:

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

In our case, we can identify:
- \( u = 1 \)
- \( v = \log(\cos x) \)

### Step 1: Differentiate \( u \) and \( v \)

1. Differentiate \( u \):
\[ \frac{du}{dx} = 0 \]

2. Differentiate \( v \):
\[ v = \log(\cos x) \]
Using the chain rule, we have:
\[ \frac{dv}{dx} = \frac{1}{\cos x} \cdot (-\sin x) = -\tan x \]

### Step 2: Apply the Quotient Rule

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

\[ \frac{dy}{dx} = \frac{(\log(\cos x))(0) - (1)(-\tan x)}{(\log(\cos x))^2} \]

This simplifies to:

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

### Final Result

Thus, the derivative of \( y = \frac{1}{\log(\cos x)} \) with respect to \( x \) is:

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

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