If ` y=(sin x)^(tan x)+ (cosx) ^(secx) ,then (dy)/(dx)`

Options

A. ` (sin x)^(tan x) (1+log (sin x) sec^(2)x) + (cos x)^(secx) secx tan x (log (cosx )+ 1)`
B. ` (sin x)^(tan x) (1-log (sin x) sec^(2)x) + (cos x)^(secx) secx tan x (log (cosx )- 1)`
C. ` (sin x)^(tan x) (1-log (sin x) sec^(2)x) + (cos x)^(secx) secx tan x (log (cosx )+ 1)`
D. ` (sin x)^(tan x) (1+log (sin x) sec^(2)x) + (cos x)^(secx) secx tan x (log (cosx )- 1)`

Answer

To differentiate the function \( y = (\sin x)^{\tan x} + (\cos x)^{\sec x} \), we will break it down into two parts and apply the logarithmic differentiation method.

### Step-by-step Solution:

1. **Identify the parts of the function**:
Let \( u = (\sin x)^{\tan x} \) and \( v = (\cos x)^{\sec x} \).
Thus, we can write:
\[ y = u + v \]

2. **Differentiate \( u \)**:
To differentiate \( u \), we will use logarithmic differentiation.
\[ \log u = \tan x \cdot \log(\sin x) \]
Now, differentiate both sides with respect to \( x \):
\[ \frac{1}{u} \frac{du}{dx} = \sec^2 x \cdot \log(\sin x) + \tan x \cdot \frac{\cos x}{\sin x} \]
Simplifying, we have:
\[ \frac{du}{dx} = u \left( \sec^2 x \cdot \log(\sin x) + \tan x \cdot \cot x \right) \]
Since \( u = (\sin x)^{\tan x} \):
\[ \frac{du}{dx} = (\sin x)^{\tan x} \left( \sec^2 x \cdot \log(\sin x) + 1 \right) \]

3. **Differentiate \( v \)**:
Similarly, for \( v \):
\[ \log v = \sec x \cdot \log(\cos x) \]
Differentiate both sides:
\[ \frac{1}{v} \frac{dv}{dx} = \sec x \cdot \left(-\tan x\right) + \sec^2 x \cdot \log(\cos x) \]
Thus,
\[ \frac{dv}{dx} = v \left( -\sec x \tan x + \sec^2 x \cdot \log(\cos x) \right) \]
Since \( v = (\cos x)^{\sec x} \):
\[ \frac{dv}{dx} = (\cos x)^{\sec x} \left( -\sec x \tan x + \sec^2 x \cdot \log(\cos x) \right) \]

4. **Combine the derivatives**:
Now we can find \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx} \]
Substitute \( \frac{du}{dx} \) and \( \frac{dv}{dx} \):
\[ \frac{dy}{dx} = (\sin x)^{\tan x} \left( \sec^2 x \cdot \log(\sin x) + 1 \right) + (\cos x)^{\sec x} \left( -\sec x \tan x + \sec^2 x \cdot \log(\cos x) \right) \]

5. **Final expression**:
Thus, the final expression for \( \frac{dy}{dx} \) is:
\[ \frac{dy}{dx} = (\sin x)^{\tan x} \left( \sec^2 x \cdot \log(\sin x) + 1 \right) + (\cos x)^{\sec x} \left( -\sec x \tan x + \sec^2 x \cdot \log(\cos x) \right) \]