If `y=10^(cot^(-1))+10^(tan^(-1)),"then "(dy)/(dx)=`

Options

A. ` (log 100)/( 1+x^(2))(10^(cot -1) -10^(tan -1_x))`
B. ` (log 100)/( 1+x^(2))(10^(tan -1_x) -10^(cot -1_x))`
C. ` (log 10)/( 1+x^(2))(10^(cot -1) -10^(tan -1_x))`
D. ` (log 10)/( 1+x^(2))(10^(tan -1_x) -10^(cot -1_x))`

Answer

To find the derivative of the function \( y = 10^{\cot^{-1}(x)} + 10^{\tan^{-1}(x)} \), we will use the chain rule and properties of logarithms. Let's go through the solution step by step.

### Step 1: Rewrite the Function
We start with:
\[ y = 10^{\cot^{-1}(x)} + 10^{\tan^{-1}(x)} \]

### Step 2: Differentiate Each Term
We will differentiate each term separately. For a function of the form \( a^{f(x)} \), the derivative is given by:
\[ \frac{d}{dx}(a^{f(x)}) = a^{f(x)} \ln(a) \cdot f'(x) \]
In our case, \( a = 10 \) and \( f(x) = \cot^{-1}(x) \) for the first term and \( f(x) = \tan^{-1}(x) \) for the second term.

### Step 3: Differentiate the First Term
For the first term \( 10^{\cot^{-1}(x)} \):
\[ \frac{d}{dx}(10^{\cot^{-1}(x)}) = 10^{\cot^{-1}(x)} \ln(10) \cdot \frac{d}{dx}(\cot^{-1}(x)) \]
The derivative of \( \cot^{-1}(x) \) is:
\[ \frac{d}{dx}(\cot^{-1}(x)) = -\frac{1}{1+x^2} \]
Thus, we have:
\[ \frac{d}{dx}(10^{\cot^{-1}(x)}) = 10^{\cot^{-1}(x)} \ln(10) \cdot \left(-\frac{1}{1+x^2}\right) \]

### Step 4: Differentiate the Second Term
For the second term \( 10^{\tan^{-1}(x)} \):
\[ \frac{d}{dx}(10^{\tan^{-1}(x)}) = 10^{\tan^{-1}(x)} \ln(10) \cdot \frac{d}{dx}(\tan^{-1}(x)) \]
The derivative of \( \tan^{-1}(x) \) is:
\[ \frac{d}{dx}(\tan^{-1}(x)) = \frac{1}{1+x^2} \]
Thus, we have:
\[ \frac{d}{dx}(10^{\tan^{-1}(x)}) = 10^{\tan^{-1}(x)} \ln(10) \cdot \frac{1}{1+x^2} \]

### Step 5: Combine the Derivatives
Now, we combine the derivatives of both terms:
\[ \frac{dy}{dx} = 10^{\cot^{-1}(x)} \ln(10) \left(-\frac{1}{1+x^2}\right) + 10^{\tan^{-1}(x)} \ln(10) \cdot \frac{1}{1+x^2} \]

### Step 6: Factor Out Common Terms
We can factor out \( \ln(10) \) and \( \frac{1}{1+x^2} \):
\[ \frac{dy}{dx} = \ln(10) \cdot \frac{1}{1+x^2} \left( 10^{\tan^{-1}(x)} - 10^{\cot^{-1}(x)} \right) \]

### Final Answer
Thus, the derivative \( \frac{dy}{dx} \) is:
\[ \frac{dy}{dx} = \frac{\ln(10)}{1+x^2} \left( 10^{\tan^{-1}(x)} - 10^{\cot^{-1}(x)} \right) \]

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