Question

Differentiate In(secx+tanx)

Answer 1

To differentiate the function \( y = \ln(\sec x + \tan x) \), we will use the chain rule and the derivatives of the functions involved.

The derivative of \( \ln(u) \) with respect to \( x \) is given by:

\[
\frac{d}{dx} \ln(u) = \frac{1}{u} \cdot \frac{du}{dx}
\]

In this case, let \( u = \sec x + \tan x \). To find the derivative, we'll first need to differentiate \( u \):

1. **Differentiate \( \sec x \)**:
\[
\frac{d}{dx}(\sec x) = \sec x \tan x
\]

2. **Differentiate \( \tan x \)**:
\[
\frac{d}{dx}(\tan x) = \sec^2 x
\]

Now, we combine these to find \( \frac{du}{dx} \):

\[
\frac{du}{dx} = \frac{d}{dx}(\sec x + \tan x) = \sec x \tan x + \sec^2 x
\]

Now we can find \( \frac{dy}{dx} \):

\[
\frac{dy}{dx} = \frac{1}{\sec x + \tan x} \cdot \left( \sec x \tan x + \sec^2 x \right)
\]

Thus, the derivative of \( y = \ln(\sec x + \tan x) \) is:

\[
\frac{dy}{dx} = \frac{\sec x \tan x + \sec^2 x}{\sec x + \tan x}
\]

This is the final result for the differentiation of \( \ln(\sec x + \tan x) \).