Differentiate the following w.r.t. x :
`(sinx)^(secx)+(tanx)^(cosx)`

Answer

To differentiate the expression \( (sin x)^{sec x} + (tan x)^{cos x} \) with respect to \( x \), we will use logarithmic differentiation for each term separately.

### Step 1: Differentiate \( (sin x)^{sec x} \)

Let \( P = (sin x)^{sec x} \).

1. Take the natural logarithm of both sides:
\[ \ln P = sec x \cdot \ln(sin x) \]

2. Differentiate both sides with respect to \( x \):
\[ \frac{1}{P} \frac{dP}{dx} = \frac{d}{dx}(sec x \cdot \ln(sin x)) \]

3. Apply the product rule on the right-hand side:
\[ \frac{d}{dx}(sec x \cdot \ln(sin x)) = \frac{d(sec x)}{dx} \cdot \ln(sin x) + sec x \cdot \frac{d(\ln(sin x))}{dx} \]

4. Calculate the derivatives:
- \( \frac{d(sec x)}{dx} = sec x \cdot tan x \)
- \( \frac{d(\ln(sin x))}{dx} = \frac{cos x}{sin x} = cot x \)

5. Substitute back:
\[ \frac{1}{P} \frac{dP}{dx} = sec x \cdot tan x \cdot \ln(sin x) + sec x \cdot cot x \]

6. Multiply through by \( P \):
\[ \frac{dP}{dx} = P \left( sec x \cdot tan x \cdot \ln(sin x) + sec x \cdot cot x \right) \]

7. Substitute back \( P = (sin x)^{sec x} \):
\[ \frac{dP}{dx} = (sin x)^{sec x} \left( sec x \cdot tan x \cdot \ln(sin x) + sec x \cdot cot x \right) \]

### Step 2: Differentiate \( (tan x)^{cos x} \)

Let \( Q = (tan x)^{cos x} \).

1. Take the natural logarithm of both sides:
\[ \ln Q = cos x \cdot \ln(tan x) \]

2. Differentiate both sides with respect to \( x \):
\[ \frac{1}{Q} \frac{dQ}{dx} = \frac{d}{dx}(cos x \cdot \ln(tan x)) \]

3. Apply the product rule:
\[ \frac{d}{dx}(cos x \cdot \ln(tan x)) = \frac{d(cos x)}{dx} \cdot \ln(tan x) + cos x \cdot \frac{d(\ln(tan x))}{dx} \]

4. Calculate the derivatives:
- \( \frac{d(cos x)}{dx} = -sin x \)
- \( \frac{d(\ln(tan x))}{dx} = \frac{sec^2 x}{tan x} \)

5. Substitute back:
\[ \frac{1}{Q} \frac{dQ}{dx} = -sin x \cdot \ln(tan x) + cos x \cdot \frac{sec^2 x}{tan x} \]

6. Multiply through by \( Q \):
\[ \frac{dQ}{dx} = Q \left( -sin x \cdot \ln(tan x) + cos x \cdot \frac{sec^2 x}{tan x} \right) \]

7. Substitute back \( Q = (tan x)^{cos x} \):
\[ \frac{dQ}{dx} = (tan x)^{cos x} \left( -sin x \cdot \ln(tan x) + cos x \cdot \frac{sec^2 x}{tan x} \right) \]

### Step 3: Combine the results

Now, we can combine the derivatives of \( P \) and \( Q \):
\[ \frac{dy}{dx} = \frac{dP}{dx} + \frac{dQ}{dx} \]
Substituting the expressions we found:
\[ \frac{dy}{dx} = (sin x)^{sec x} \left( sec x \cdot tan x \cdot \ln(sin x) + sec x \cdot cot x \right) + (tan x)^{cos x} \left( -sin x \cdot \ln(tan x) + cos x \cdot \frac{sec^2 x}{tan x} \right) \]

### Final Answer
Thus, the derivative of \( (sin x)^{sec x} + (tan x)^{cos x} \) with respect to \( x \) is:
\[ \frac{dy}{dx} = (sin x)^{sec x} \left( sec x \cdot tan x \cdot \ln(sin x) + sec x \cdot cot x \right) + (tan x)^{cos x} \left( -sin x \cdot \ln(tan x) + cos x \cdot \frac{sec^2 x}{tan x} \right) \]