`y=(tanx)^(logx)+(cosx)^(sinx)`, find `dy/dx`

Answer

To find the derivative \( \frac{dy}{dx} \) for the function

\[ y = (\tan x)^{\log x} + (\cos x)^{\sin x}, \]

we will differentiate each term separately.

### Step 1: Differentiate the first term \( u = (\tan x)^{\log x} \)

1. **Take the logarithm of both sides**:
\[ \log u = \log((\tan x)^{\log x}) = \log x \cdot \log(\tan x). \]

2. **Differentiate both sides using implicit differentiation**:
\[ \frac{1}{u} \frac{du}{dx} = \frac{d}{dx}(\log x \cdot \log(\tan x)). \]

3. **Apply the product rule**:
\[ \frac{d}{dx}(\log x \cdot \log(\tan x)) = \log(\tan x) \cdot \frac{1}{x} + \log x \cdot \frac{1}{\tan x} \cdot \sec^2 x. \]

4. **Multiply both sides by \( u \)** to solve for \( \frac{du}{dx} \):
\[ \frac{du}{dx} = u \left( \frac{\log(\tan x)}{x} + \frac{\log x \cdot \sec^2 x}{\tan x} \right). \]

5. **Substitute back \( u = (\tan x)^{\log x} \)**:
\[ \frac{du}{dx} = (\tan x)^{\log x} \left( \frac{\log(\tan x)}{x} + \frac{\log x \cdot \sec^2 x}{\tan x} \right). \]

### Step 2: Differentiate the second term \( v = (\cos x)^{\sin x} \)

1. **Take the logarithm of both sides**:
\[ \log v = \sin x \cdot \log(\cos x). \]

2. **Differentiate both sides using implicit differentiation**:
\[ \frac{1}{v} \frac{dv}{dx} = \cos x \cdot \log(\cos x) + \sin x \cdot \frac{-\sin x}{\cos x}. \]

3. **Multiply both sides by \( v \)** to solve for \( \frac{dv}{dx} \):
\[ \frac{dv}{dx} = v \left( \log(\cos x) \cdot \cos x - \sin^2 x \cdot \frac{1}{\cos x} \right). \]

4. **Substitute back \( v = (\cos x)^{\sin x} \)**:
\[ \frac{dv}{dx} = (\cos x)^{\sin x} \left( \log(\cos x) \cdot \cos x - \frac{\sin^2 x}{\cos x} \right). \]

### Step 3: Combine the derivatives

Now that we have both derivatives, we can find \( \frac{dy}{dx} \):

\[ \frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx}. \]

Substituting the expressions we found:

\[ \frac{dy}{dx} = (\tan x)^{\log x} \left( \frac{\log(\tan x)}{x} + \frac{\log x \cdot \sec^2 x}{\tan x} \right) + (\cos x)^{\sin x} \left( \log(\cos x) \cdot \cos x - \frac{\sin^2 x}{\cos x} \right). \]

### Final Result

Thus, the derivative \( \frac{dy}{dx} \) is:

\[ \frac{dy}{dx} = (\tan x)^{\log x} \left( \frac{\log(\tan x)}{x} + \frac{\log x \cdot \sec^2 x}{\tan x} \right) + (\cos x)^{\sin x} \left( \log(\cos x) \cdot \cos x - \frac{\sin^2 x}{\cos x} \right). \]

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