The derivative of ` a^(sec x ) " w.r.t."a^(tan x ) (a gt 0) ` is

Options

A. `sec x a^(secx - tanx)`
B. ` sin x a^(tan x - sec x )`
C. `sin x a^(sec x - tan x)`
D. `a ^(sec x - tan x)`

Explanation

To find the derivative of \( a^{\sec x} \) with respect to \( a^{\tan x} \) (where \( a > 0 \)), we can follow these steps:

### Step-by-Step Solution:

1. **Define the Variables**:
Let \( U = a^{\sec x} \) and \( V = a^{\tan x} \).

2. **Take the Natural Logarithm**:
To differentiate, we can use logarithmic differentiation. We take the natural logarithm of both \( U \) and \( V \):
\[ \ln U = \sec x \ln a \]
\[ \ln V = \tan x \ln a \]

3. **Differentiate Both Sides**:
Now, we differentiate both sides with respect to \( x \):
\[ \frac{dU}{dx} = U \cdot \frac{d}{dx}(\sec x \ln a) = a^{\sec x} \cdot \ln a \cdot \sec x \tan x \]
\[ \frac{dV}{dx} = V \cdot \frac{d}{dx}(\tan x \ln a) = a^{\tan x} \cdot \ln a \cdot \sec^2 x \]

4. **Find the Derivative \( \frac{dU}{dV} \)**:
Using the chain rule, we find \( \frac{dU}{dV} \):
\[ \frac{dU}{dV} = \frac{\frac{dU}{dx}}{\frac{dV}{dx}} = \frac{a^{\sec x} \cdot \ln a \cdot \sec x \tan x}{a^{\tan x} \cdot \ln a \cdot \sec^2 x} \]

5. **Simplify the Expression**:
We can simplify this expression:
\[ \frac{dU}{dV} = \frac{a^{\sec x}}{a^{\tan x}} \cdot \frac{\sec x \tan x}{\sec^2 x} = a^{\sec x - \tan x} \cdot \frac{\tan x}{\sec x} \]
\[ = a^{\sec x - \tan x} \cdot \sin x \]

6. **Final Result**:
Thus, the derivative of \( a^{\sec x} \) with respect to \( a^{\tan x} \) is:
\[ \frac{dU}{dV} = a^{\sec x - \tan x} \cdot \sin x \]