If `y=(sinx)^(tanx),then(dy)/(dx)` is equal to

Options

A. `(sinx)^(tanx).[1+sec^2x.log(sinx)]`
B. `tanx.(sinx)^(tanx-1).cosx`
C. `(sinx)^(tanx).sec^2.x.log(sinx)`
D. `tanx.(sinx)^(tanx-1)`

Answer

To find the derivative of the function \( y = (\sin x)^{\tan x} \), we can use logarithmic differentiation. Here’s a step-by-step solution:

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

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

### Step 3: Differentiate both sides with respect to \( x \)
Now, we differentiate both sides with respect to \( x \). Remember to use the product rule on the right side:
\[ \frac{d}{dx}(\ln y) = \frac{d}{dx}(\tan x \cdot \ln(\sin x)) \]

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

### Step 4: Differentiate the components
Now we differentiate the components:
1. The derivative of \( \tan x \) is \( \sec^2 x \).
2. The derivative of \( \ln(\sin x) \) is \( \frac{\cos x}{\sin x} = \cot x \).

Thus, we have:
\[ \frac{1}{y} \frac{dy}{dx} = \sec^2 x \cdot \ln(\sin x) + \tan x \cdot \cot x \]

### Step 5: Simplify the expression
Since \( \tan x \cdot \cot x = 1 \), we can simplify:
\[ \frac{1}{y} \frac{dy}{dx} = \sec^2 x \cdot \ln(\sin x) + 1 \]

### Step 6: Solve for \( \frac{dy}{dx} \)
Now, multiply both sides by \( y \) to isolate \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = y \left( \sec^2 x \cdot \ln(\sin x) + 1 \right) \]

### Step 7: Substitute back for \( y \)
Recall that \( y = (\sin x)^{\tan x} \):
\[ \frac{dy}{dx} = (\sin x)^{\tan x} \left( \sec^2 x \cdot \ln(\sin x) + 1 \right) \]

### Final Result
Thus, the derivative \( \frac{dy}{dx} \) is:
\[ \frac{dy}{dx} = (\sin x)^{\tan x} \left( \sec^2 x \cdot \ln(\sin x) + 1 \right) \]

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