If `x = e^(tan^(-1))((y-x^2)/x^2)` then `(dy)/(dx)=`

Options

A. `2x{1+tan(log_(e)x)}+xsec^(2)(log_(e)x)`
B. `x{1+tan(log_(e)x)}+sec^(2)(log_(e)x)`
C. `2x{1+tan(log_(e)x)}+x^(2)sec^(2)(log_(e)x)`
D. `2x{1+tan(log_(e)x)}+sec^(2)(log_(e)x)`

Answer

To solve the problem where \( x = e^{\tan^{-1}\left(\frac{y - x^2}{x^2}\right)} \) and we need to find \( \frac{dy}{dx} \), we can follow these steps:

### Step 1: Take the natural logarithm of both sides
We start with the equation:
\[ x = e^{\tan^{-1}\left(\frac{y - x^2}{x^2}\right)} \]
Taking the natural logarithm on both sides gives:
\[ \ln x = \tan^{-1}\left(\frac{y - x^2}{x^2}\right) \]

### Step 2: Apply the tangent function
To eliminate the arctangent, we apply the tangent function to both sides:
\[ \tan(\ln x) = \frac{y - x^2}{x^2} \]

### Step 3: Rearranging the equation
Rearranging the equation to solve for \( y \):
\[ y - x^2 = x^2 \tan(\ln x) \]
Thus,
\[ y = x^2 \tan(\ln x) + x^2 \]

### Step 4: Differentiate both sides with respect to \( x \)
Now we differentiate \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = \frac{d}{dx}[x^2 \tan(\ln x) + x^2] \]
Using the product rule on the first term \( x^2 \tan(\ln x) \):
\[ \frac{dy}{dx} = \frac{d}{dx}[x^2] \cdot \tan(\ln x) + x^2 \cdot \frac{d}{dx}[\tan(\ln x)] + \frac{d}{dx}[x^2] \]

### Step 5: Calculate the derivatives
1. The derivative of \( x^2 \) is \( 2x \).
2. For \( \tan(\ln x) \), we use the chain rule:
\[ \frac{d}{dx}[\tan(\ln x)] = \sec^2(\ln x) \cdot \frac{d}{dx}[\ln x] = \sec^2(\ln x) \cdot \frac{1}{x} \]

### Step 6: Substitute back into the derivative equation
Substituting these derivatives back into our equation:
\[ \frac{dy}{dx} = 2x \tan(\ln x) + x^2 \cdot \sec^2(\ln x) \cdot \frac{1}{x} + 2x \]
This simplifies to:
\[ \frac{dy}{dx} = 2x \tan(\ln x) + x \sec^2(\ln x) + 2x \]

### Step 7: Factor the expression
We can factor out \( 2x \) from the first and last terms:
\[ \frac{dy}{dx} = 2x(1 + \tan(\ln x)) + x \sec^2(\ln x) \]

### Final Result
Thus, the final expression for \( \frac{dy}{dx} \) is:
\[ \frac{dy}{dx} = 2x(1 + \tan(\ln x)) + x \sec^2(\ln x) \]

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