The derivative of `a^(secx)` w.r.t. `a^(tanx)(agt0)` is

Options

A. `secxa^(secx-tanx)`
B. `sinxa^(tanx-secx)`
C. `sinxa^(secx-tanx)`
D. `a(secx-tanx)`

Explanation

To find the derivative of \( a^{\sec x} \) with respect to \( a^{\tan x} \), we will use the chain rule and the properties of derivatives. Let's denote:

- \( u = a^{\sec x} \)
- \( v = a^{\tan x} \)

We want to find \( \frac{du}{dv} \). By the chain rule, we can express this as:

\[ \frac{du}{dv} = \frac{du/dx}{dv/dx} \]

### Step 1: Calculate \( \frac{du}{dx} \)

Using the formula for the derivative of \( a^x \), which is \( a^x \ln(a) \cdot \frac{dx}{dx} \), we have:

\[ \frac{du}{dx} = a^{\sec x} \ln(a) \cdot \frac{d}{dx}(\sec x) \]

Now, we need to find \( \frac{d}{dx}(\sec x) \):

\[ \frac{d}{dx}(\sec x) = \sec x \tan x \]

Thus,

\[ \frac{du}{dx} = a^{\sec x} \ln(a) \cdot \sec x \tan x \]

### Step 2: Calculate \( \frac{dv}{dx} \)

Similarly, for \( v = a^{\tan x} \):

\[ \frac{dv}{dx} = a^{\tan x} \ln(a) \cdot \frac{d}{dx}(\tan x) \]

Now, we need to find \( \frac{d}{dx}(\tan x) \):

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

Thus,

\[ \frac{dv}{dx} = a^{\tan x} \ln(a) \cdot \sec^2 x \]

### Step 3: Compute \( \frac{du}{dv} \)

Now we can substitute \( \frac{du}{dx} \) and \( \frac{dv}{dx} \) into the expression for \( \frac{du}{dv} \):

\[ \frac{du}{dv} = \frac{a^{\sec x} \ln(a) \sec x \tan x}{a^{\tan x} \ln(a) \sec^2 x} \]

Notice that \( \ln(a) \) cancels out:

\[ \frac{du}{dv} = \frac{a^{\sec x} \sec x \tan x}{a^{\tan x} \sec^2 x} \]

### Step 4: Simplify the expression

We can simplify further:

\[ \frac{du}{dv} = \frac{a^{\sec x}}{a^{\tan x}} \cdot \frac{\sec x \tan x}{\sec^2 x} \]

This simplifies to:

\[ \frac{du}{dv} = a^{\sec x - \tan x} \cdot \frac{\tan x}{\sec x} \]

Since \( \frac{\tan x}{\sec x} = \sin x \):

\[ \frac{du}{dv} = a^{\sec x - \tan x} \sin x \]

### Final Answer

Thus, the derivative of \( a^{\sec x} \) with respect to \( a^{\tan x} \) is:

\[ \frac{du}{dv} = a^{\sec x - \tan x} \sin x \]