Question

\( -\tan\left(\frac{1}{x}\right) + \frac{1}{x} + c = ? \)

Answer 1

Derivative of the given function:
We are asked to find the derivative of the function: \[ f(x) = -\tan\left(\frac{1}{x}\right) + \frac{1}{x} + c \]

Step 1: Apply the chain rule:
The derivative of \( -\tan\left(\frac{1}{x}\right) \) is: \[ \frac{d}{dx} \left[-\tan\left(\frac{1}{x}\right)\right] = -\sec^2\left(\frac{1}{x}\right) \cdot \left(-\frac{1}{x^2}\right) \] The derivative of \( \frac{1}{x} \) is: \[ \frac{d}{dx} \left(\frac{1}{x}\right) = -\frac{1}{x^2} \] The derivative of the constant \( c \) is 0.

Step 2: Combine the results:
Now, combining all the terms, we get: \[ \frac{d}{dx} \left[-\tan\left(\frac{1}{x}\right) + \frac{1}{x} + c\right] = \sec^2\left(\frac{1}{x}\right) \cdot \frac{1}{x^2} - \frac{1}{x^2} \] Simplifying the expression: \[ = \frac{\sec^2\left(\frac{1}{x}\right) - 1}{x^2} \] Using the identity \( \sec^2(\theta) - 1 = \tan^2(\theta) \), we get: \[ = \frac{\tan^2\left(\frac{1}{x}\right)}{x^2} \]

Final Answer:
Therefore, the derivative is: \[ \frac{\tan^2\left(\frac{1}{x}\right)}{x^2} \]