Question

If \[ x=e^{\tan^{-1}\left(\frac{y-x^2}{x}\right)} \] then \(\dfrac{dy}{dx}=?\)

• A. \(\sec^2(\log_e x)+\tan(\log_e x)\)
• B. \(2x+\sec^2(\log_e x)+\tan(\log_e x)\)
• C. \(-2x+\sec^2(\log_e x)+\tan(\log_e x)\)
• D. \(2x+\sec(\log_e x)+\tan(\log_e x)\)

Hint:

For equations involving \(e^{\tan^{-1}(\cdot)}\), first take \(\log_e\) on both sides and then apply tangent to remove \(\tan^{-1}\).

Answer 1

The Correct Option is B

Concept:
When a variable is given inside inverse trigonometric and exponential functions, first simplify the equation using logarithm and trigonometric inverse properties.

Step 1: Take natural logarithm on both sides. \[ x=e^{\tan^{-1}\left(\frac{y-x^2}{x}\right)} \] Taking \(\log_e\) on both sides: \[ \log_e x=\tan^{-1}\left(\frac{y-x^2}{x}\right) \]

Step 2: Apply tangent on both sides. \[ \tan(\log_e x)=\frac{y-x^2}{x} \]

Step 3: Solve for \(y\). \[ x\tan(\log_e x)=y-x^2 \] \[ y=x^2+x\tan(\log_e x) \]

Step 4: Differentiate both sides with respect to \(x\). \[ \frac{dy}{dx}=\frac{d}{dx}(x^2)+\frac{d}{dx}\left[x\tan(\log_e x)\right] \] \[ \frac{d}{dx}(x^2)=2x \] Now apply product rule: \[ \frac{d}{dx}\left[x\tan(\log_e x)\right] =x\cdot \frac{d}{dx}\left[\tan(\log_e x)\right]+\tan(\log_e x)\cdot \frac{d}{dx}(x) \] \[ =x\cdot \sec^2(\log_e x)\cdot \frac{1}{x}+\tan(\log_e x) \] \[ =\sec^2(\log_e x)+\tan(\log_e x) \]

Step 5: Combine the terms. \[ \frac{dy}{dx}=2x+\sec^2(\log_e x)+\tan(\log_e x) \] \[ \therefore \text{Correct Answer is (B)} \]