Differentiation

To find the derivative of with respect to at , we perform the following steps:

1. Let and .
2. We need to find , using the chain rule: .
3. First, calculate :
   • The expression for is: .
   • Let , then .
   • By chain rule, .
   • For , use quotient rule:
     • where and .
     • The derivative of with respect to is , and .
     • Thus, .
4. Similarly, calculate :
   • The expression for is: .
   • Let , then .
   • By chain rule, .
   • Differentiating = \frac{2x\sqrt{1-x^2}}{1-2x^2}\) using quotient rule:
     • }\) , whe\)
     • The derivative of the numerator is
     • Applying product rule to :
5. Evaluate each derivative at .
6. Plug the values into and find the ratio at .

After calculating, the derivative evaluates to at .

To solve the problem, we need to find the equation of another tangent to a circle from the origin. Given that the line is a tangent to the circle centered at , we can use the property that the perpendicular from the center of the circle to the tangent line will also be the radius of the circle.

The given tangent line to the circle is . This line has a slope .

The center of the circle is .

Using the point-to-line distance formula, the perpendicular distance ( ) from the center to the line is given by:

Applying the values to the distance formula:

The other tangent from the origin will also make the same angle with the origin-center line. The original line is having slope . Lines from the same point (origin here) having equal angles to the same line will have slopes related as negative reciprocals, so we need a line tangent from origin.

Since the slopes of the tangents are related through negative reciprocals (for perpendicular tangents), the corresponding slope can either be or . However, we want the equation of another tangent , which corresponds to the slope being .

Hence, the equation of the other tangent from the origin is .