If : `f'(1)=2" and "g'(sqrt(2))=4," then: derivative of "f(tanx)w.r.t.g(secx)" at "x=pi//4,` is

Options

A. `(1)/(sqrt(2))`
B. `sqrt(2)`
C. 1
D. `(3)/(2)`

Answer

To find the derivative of \( f(\tan x) \) with respect to \( g(\sec x) \) at \( x = \frac{\pi}{4} \), we will use the chain rule and the given information about the derivatives of \( f \) and \( g \).

### Step-by-Step Solution:

1. **Define the Functions:**
Let \( u = f(\tan x) \) and \( v = g(\sec x) \).

2. **Find \( \frac{du}{dx} \):**
Using the chain rule:
\[ \frac{du}{dx} = f'(\tan x) \cdot \frac{d}{dx}(\tan x) \]
The derivative of \( \tan x \) is \( \sec^2 x \). Thus,
\[ \frac{du}{dx} = f'(\tan x) \cdot \sec^2 x \]

3. **Find \( \frac{dv}{dx} \):**
Similarly, for \( v \):
\[ \frac{dv}{dx} = g'(\sec x) \cdot \frac{d}{dx}(\sec x) \]
The derivative of \( \sec x \) is \( \sec x \tan x \). Therefore,
\[ \frac{dv}{dx} = g'(\sec x) \cdot \sec x \tan x \]

4. **Find \( \frac{du}{dv} \):**
We want to find \( \frac{du}{dv} \):
\[ \frac{du}{dv} = \frac{\frac{du}{dx}}{\frac{dv}{dx}} = \frac{f'(\tan x) \cdot \sec^2 x}{g'(\sec x) \cdot \sec x \tan x} \]

5. **Evaluate at \( x = \frac{\pi}{4} \):**
At \( x = \frac{\pi}{4} \):
- \( \tan\left(\frac{\pi}{4}\right) = 1 \)
- \( \sec\left(\frac{\pi}{4}\right) = \sqrt{2} \)

Substitute these values:
\[ \frac{du}{dv} = \frac{f'(1) \cdot \sec^2\left(\frac{\pi}{4}\right)}{g'(\sqrt{2}) \cdot \sec\left(\frac{\pi}{4}\right) \tan\left(\frac{\pi}{4}\right)} \]

We know:
- \( \sec^2\left(\frac{\pi}{4}\right) = 2 \)
- \( f'(1) = 2 \)
- \( g'(\sqrt{2}) = 4 \)

Thus,
\[ \frac{du}{dv} = \frac{2 \cdot 2}{4 \cdot \sqrt{2} \cdot 1} = \frac{4}{4\sqrt{2}} = \frac{1}{\sqrt{2}} \]

### Final Answer:
The derivative of \( f(\tan x) \) with respect to \( g(\sec x) \) at \( x = \frac{\pi}{4} \) is:
\[ \frac{1}{\sqrt{2}} \]