If x = exp {tan-1(y−x²/x²)}, then dy/dx equals

The correct option is: 2x [1 + tan (log x)] + x sec² (log x)

Step-by-step derivation:

We are given the relation:
$$x=\exp \left\{{\tan }^{-1}\left(\frac{y-x^2}{x^2}\right)\right\}$$

Taking the natural logarithm ($\log$) on both sides of the equation:
$$\log x={\tan }^{-1}\left(\frac{y-x^2}{x^2}\right)$$

Taking the tangent ($\tan$) on both sides:
$$\tan (\log x)=\frac{y-x^2}{x^2}$$

Now, we solve for $y$ by multiplying both sides by $x^2$:
$$x^2\tan (\log x)=y-x^2$$
Adding $x^2$ to both sides gives:
$$y=x^2+x^2\tan (\log x)$$
Factoring out $x^2$:
$$y=x^2[1+\tan (\log x\left)\right]$$

Now, we differentiate $y$ with respect to $x$ using the product rule:
$$\frac{dy}{dx}=\frac{d}{dx}\left(x^2\right)\cdot [1+\tan (\log x\left)\right]+x^2\cdot \frac{d}{dx}[1+\tan (\log x\left)\right]$$

Compute the individual derivatives:
1) The derivative of the first term is:
$$\frac{d}{dx}\left(x^2\right)=2x$$
2) The derivative of the second term using the chain rule is:
$$\frac{d}{dx}[1+\tan (\log x\left)\right]={\sec }^2(\log x)\cdot \frac{1}{x}$$

Substituting these derivatives back into the product rule formula:
$$\frac{dy}{dx}=2x[1+\tan (\log x\left)\right]+x^2\cdot [{\sec }^2(\log x)\cdot \frac{1}{x}]$$
Simplifying the second term by cancelling out $x$:
$$\frac{dy}{dx}=2x[1+\tan (\log x\left)\right]+x{\sec }^2(\log x)$$