Find `(dy)/(dx)`, when:
`y=(tanx)^(sinx)`

Answer

To find \(\frac{dy}{dx}\) where \(y = (\tan x)^{\sin x}\), we will use logarithmic differentiation. Here’s a step-by-step solution:

### Step 1: Take the natural logarithm of both sides
We start by taking the logarithm of both sides to simplify the differentiation:
\[ \ln y = \ln((\tan x)^{\sin x}) \]

### Step 2: Apply the logarithmic property
Using the property of logarithms that states \(\ln(a^b) = b \ln a\), we can rewrite the equation:
\[ \ln y = \sin x \cdot \ln(\tan x) \]

### Step 3: Differentiate both sides
Now, we differentiate both sides with respect to \(x\):
\[ \frac{d}{dx}(\ln y) = \frac{d}{dx}(\sin x \cdot \ln(\tan x)) \]

Using the chain rule on the left side:
\[ \frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}(\sin x) \cdot \ln(\tan x) + \sin x \cdot \frac{d}{dx}(\ln(\tan x)) \]

### Step 4: Differentiate the right side using the product rule
Now we apply the product rule on the right side:
\[ \frac{d}{dx}(\sin x) = \cos x \]
For \(\frac{d}{dx}(\ln(\tan x))\), we use the chain rule:
\[ \frac{d}{dx}(\ln(\tan x)) = \frac{1}{\tan x} \cdot \sec^2 x = \frac{\sec^2 x}{\tan x} \]

### Step 5: Substitute back into the equation
Now substituting these derivatives back into our equation:
\[ \frac{1}{y} \frac{dy}{dx} = \cos x \cdot \ln(\tan x) + \sin x \cdot \frac{\sec^2 x}{\tan x} \]

### Step 6: Multiply both sides by \(y\)
To isolate \(\frac{dy}{dx}\), we multiply both sides by \(y\):
\[ \frac{dy}{dx} = y \left( \cos x \cdot \ln(\tan x) + \sin x \cdot \frac{\sec^2 x}{\tan x} \right) \]

### Step 7: Substitute \(y\) back in
Recall that \(y = (\tan x)^{\sin x}\):
\[ \frac{dy}{dx} = (\tan x)^{\sin x} \left( \cos x \cdot \ln(\tan x) + \sin x \cdot \frac{\sec^2 x}{\tan x} \right) \]

### Final Answer
Thus, the final answer is:
\[ \frac{dy}{dx} = (\tan x)^{\sin x} \left( \cos x \cdot \ln(\tan x) + \sin x \cdot \frac{\sec^2 x}{\tan x} \right) \]

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