If ` v =  log ( 1 + x^(2)) and u = x - tan^(-1) x ` then ` ,( du)/( dv)` is equal to

Options

A. `e ^(x) - y `
B. ` e^(x) - 1 `
C. 2/x
D. ` x^(2) - 1 `

Answer

To find \(\frac{du}{dv}\), we will first need to compute \(\frac{du}{dx}\) and \(\frac{dv}{dx}\), and then use the relationship \(\frac{du}{dv} = \frac{du/dx}{dv/dx}\).

### Step 1: Calculate \(v = \log(1 + x^2)\)

To find \(\frac{dv}{dx}\):
\[ v = \log(1 + x^2) \]
Using the chain rule:
\[ \frac{dv}{dx} = \frac{1}{1 + x^2} \cdot \frac{d}{dx}(1 + x^2) \]
The derivative of \(1 + x^2\) is:
\[ \frac{d}{dx}(1 + x^2) = 0 + 2x = 2x \]
Thus,
\[ \frac{dv}{dx} = \frac{1}{1 + x^2} \cdot 2x = \frac{2x}{1 + x^2} \]

### Step 2: Calculate \(u = x - \tan^{-1}(x)\)

Now, we find \(\frac{du}{dx}\):
\[ u = x - \tan^{-1}(x) \]
The derivative of \(x\) is \(1\) and the derivative of \(\tan^{-1}(x)\) is \(\frac{1}{1 + x^2}\):
\[ \frac{du}{dx} = 1 - \frac{1}{1 + x^2} \]
Simplifying this:
\[ \frac{du}{dx} = 1 - \frac{1}{1 + x^2} = \frac{(1 + x^2) - 1}{1 + x^2} = \frac{x^2}{1 + x^2} \]

### Step 3: Find \(\frac{du}{dv}\)

Now we can find \(\frac{du}{dv}\):
\[ \frac{du}{dv} = \frac{du/dx}{dv/dx} = \frac{\frac{x^2}{1 + x^2}}{\frac{2x}{1 + x^2}} \]
The \((1 + x^2)\) terms cancel out:
\[ \frac{du}{dv} = \frac{x^2}{2x} \]
This simplifies to:
\[ \frac{du}{dv} = \frac{x}{2} \]

### Final Answer
Thus, the value of \(\frac{du}{dv}\) is:
\[ \frac{du}{dv} = \frac{x}{2} \]