Answer is: option1

\( y + \frac{\pi}{3} = 10x + \sqrt{3} \)

Solution:

Finding the Equation of the Tangent Line

To find the slope of the tangent line, we first differentiate \[ y = \sin^{-1}(5x) \] using the chain rule:

\[ \frac{d}{dx} \sin^{-1}(u) = \frac{1}{\sqrt{1 - u^2}} \cdot \frac{du}{dx} \]

where \( u = 5x \), so \( \frac{du}{dx} = 5 \). Therefore, the derivative is:

\[ \frac{dy}{dx} = \frac{5}{\sqrt{1 - (5x)^2}} \]

Substituting \( x = -\frac{\sqrt{3}}{10} \):

\[ 1 - (5x)^2 = 1 - 25 \left(\frac{3}{100}\right) = 1 - \frac{75}{100} = 1 - 0.75 = 0.25 \] \[ \sqrt{0.25} = \frac{1}{2} \] \[ \frac{dy}{dx} = \frac{5}{\frac{1}{2}} = 10 \]

So, the slope of the tangent line is \( 10 \).

Finding the Point on the Curve

We substitute \( x = -\frac{\sqrt{3}}{10} \) into the original function:

\[ y = \sin^{-1} \left( 5 \times -\frac{\sqrt{3}}{10} \right) \] \[ y = \sin^{-1} \left(-\frac{\sqrt{3}}{2} \right) \]

From trigonometric identities:

\[ \sin^{-1} \left(-\frac{\sqrt{3}}{2} \right) = -\frac{\pi}{3} \]

So, the point on the curve is:

\[ \left( -\frac{\sqrt{3}}{10}, -\frac{\pi}{3} \right) \] Equation of the Tangent Line

The equation of a tangent line is given by:

\[ y - y_1 = m(x - x_1) \]

Substituting \( y_1 = -\frac{\pi}{3} \), \( m = 10 \), and \( x_1 = -\frac{\sqrt{3}}{10} \):

\[ y + \frac{\pi}{3} = 10 \left( x + \frac{\sqrt{3}}{10} \right) \] \[ y + \frac{\pi}{3} = 10x + \sqrt{3} \] Final Answer:

The correct equation of the tangent line is:

\[ y + \frac{\pi}{3} = 10x + \sqrt{3} \]