If ` y= ( 10)^((10) ^(log _10 ^(cosec (10x)))),then (dy)/(dx) =`

Options

A. ` -10 cosec (10x) cot (10x) (log10) 10 ^(cosec (10x))`
B. ` 10 cosec (10x) cot (10x) (log10) 10 ^(cosec (10x))`
C. `- 10 cosec (10x) cot (10x) (log10)^(2) 10 ^(cosec (10x))`
D. ` 10 cosec (10x) cot (10x) (log10)^(2) 10 ^(cosec (10x))`

Answer

To solve the problem \( y = 10^{10^{\log_{10}(\csc(10x))}} \) and find \( \frac{dy}{dx} \), we will follow these steps:

### Step 1: Simplify the expression for \( y \)

We can use the property of logarithms that states \( a^{\log_b(c)} = c^{\log_b(a)} \). Applying this property, we can rewrite \( y \):

\[ y = 10^{10^{\log_{10}(\csc(10x))}} = 10^{\csc(10x)^{\log_{10}(10)}} \]

Since \( \log_{10}(10) = 1 \), we have:

\[ y = 10^{\csc(10x)} \]

### Step 2: Differentiate \( y \)

To differentiate \( y \), we will use the formula for the derivative of an exponential function \( a^{f(x)} \):

\[ \frac{d}{dx}(a^{f(x)}) = a^{f(x)} \cdot \ln(a) \cdot f'(x) \]

In our case, \( a = 10 \) and \( f(x) = \csc(10x) \). Thus, we have:

\[ \frac{dy}{dx} = 10^{\csc(10x)} \cdot \ln(10) \cdot \frac{d}{dx}(\csc(10x)) \]

### Step 3: Differentiate \( \csc(10x) \)

The derivative of \( \csc(x) \) is:

\[ \frac{d}{dx}(\csc(x)) = -\csc(x) \cot(x) \]

Using the chain rule, we differentiate \( \csc(10x) \):

\[ \frac{d}{dx}(\csc(10x)) = -\csc(10x) \cot(10x) \cdot \frac{d}{dx}(10x) = -10 \csc(10x) \cot(10x) \]

### Step 4: Substitute back into the derivative

Now substituting back into our derivative expression:

\[ \frac{dy}{dx} = 10^{\csc(10x)} \cdot \ln(10) \cdot (-10 \csc(10x) \cot(10x)) \]

This simplifies to:

\[ \frac{dy}{dx} = -10^{\csc(10x) + 1} \cdot \ln(10) \cdot \csc(10x) \cot(10x) \]

### Final Answer

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

\[ \frac{dy}{dx} = -10^{\csc(10x) + 1} \ln(10) \csc(10x) \cot(10x) \]