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If x = exp {tan^(-1)((y-x^2)/(x^2))} the...

If `x = exp {tan^(-1)((y-x^2)/(x^2))}` then dy/dx equals

A

`2x[1+tan(logx)]+x sec^2(logx)`

B

`x[1+tan(logx)]+x sec^2(logx)`

C

`2x[1+tan(logx)]+x^2 sec^2(logx)`

D

`2x[1+tan(logx)]+ sec^2(logx)`

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To solve the equation \( x = e^{\tan^{-1}\left(\frac{y - x^2}{x^2}\right)} \) and find \( \frac{dy}{dx} \), we will 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 (ln) of 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 express \( y \): \[ y - x^2 = x^2 \tan(\ln x) \] \[ y = x^2 + x^2 \tan(\ln x) \] ### Step 4: Differentiate both sides with respect to \( x \) Now, we differentiate \( y \) with respect to \( x \): \[ \frac{dy}{dx} = \frac{d}{dx}(x^2) + \frac{d}{dx}(x^2 \tan(\ln x)) \] ### Step 5: Differentiate the first term The derivative of \( x^2 \) is: \[ \frac{d}{dx}(x^2) = 2x \] ### Step 6: Differentiate the second term using the product rule For the term \( x^2 \tan(\ln x) \), we use the product rule: \[ \frac{d}{dx}(x^2 \tan(\ln x)) = \frac{d}{dx}(x^2) \cdot \tan(\ln x) + x^2 \cdot \frac{d}{dx}(\tan(\ln x)) \] Calculating the first part: \[ \frac{d}{dx}(x^2) \cdot \tan(\ln x) = 2x \tan(\ln x) \] Now for the second part, we apply the chain rule: \[ \frac{d}{dx}(\tan(\ln x)) = \sec^2(\ln x) \cdot \frac{d}{dx}(\ln x) = \sec^2(\ln x) \cdot \frac{1}{x} \] Thus, \[ x^2 \cdot \frac{d}{dx}(\tan(\ln x)) = x^2 \cdot \sec^2(\ln x) \cdot \frac{1}{x} = x \sec^2(\ln x) \] ### Step 7: Combine the derivatives Now, we combine the derivatives: \[ \frac{dy}{dx} = 2x + 2x \tan(\ln x) + x \sec^2(\ln x) \] ### Final Expression Thus, the final expression for \( \frac{dy}{dx} \) is: \[ \frac{dy}{dx} = 2x(1 + \tan(\ln x)) + x \sec^2(\ln x) \]

To solve the equation \( x = e^{\tan^{-1}\left(\frac{y - x^2}{x^2}\right)} \) and find \( \frac{dy}{dx} \), we will 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 (ln) of both sides gives: ...
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