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To find the equation of the tangent line to the graph of \( y = \sin^{-1} \left( -\frac{x}{4} \right) \) at the point \( \left( 2, -\frac{\pi}{6} \right) \), we need to:

1. **Differentiate \( y = \sin^{-1} \left( -\frac{x}{4} \right) \) with respect to \( x \)** to get \( \frac{dy}{dx} \), the slope of the tangent line.
2. **Evaluate \( \frac{dy}{dx} \) at \( x = 2 \)** to find the slope at that point.
3. **Use the point-slope form of a line** with the given point \( \left( 2, -\frac{\pi}{6} \right) \) to write the equation of the tangent line.

### Step 1: Differentiate \( y = \sin^{-1} \left( -\frac{x}{4} \right) \)

Let \( u = -\frac{x}{4} \). Then \( y = \sin^{-1}(u) \), and we need to apply the chain rule:
\[
\frac{dy}{dx} = \frac{d}{dx} \left( \sin^{-1} (u) \right) = \frac{1}{\sqrt{1 - u^2}} \cdot \frac{du}{dx}
\]

Since \( u = -\frac{x}{4} \),
\[
\frac{du}{dx} = -\frac{1}{4}
\]

Thus,
\[
\frac{dy}{dx} = \frac{-1/4}{\sqrt{1 - \left(-\frac{x}{4}\right)^2}} = \frac{-1/4}{\sqrt{1 - \frac{x^2}{16}}}
\]

### Step 2: Evaluate \( \frac{dy}{dx} \) at \( x = 2 \)

Substitute \( x = 2 \) into the derivative:
\[
\frac{dy}{dx} \Big|_{x=2} = \frac{-1/4}{\sqrt{1 - \frac{2^2}{16}}} = \frac{-1/4}{\sqrt{1 - \frac{4}{16}}} = \frac{-1/4}{\sqrt{\frac{3}{4}}} = \frac{-1/4}{\frac{\sqrt{3}}{2}} = -\frac{1}{4} \cdot \frac{2}{\sqrt{3}} = -\frac{1}{2\sqrt{3}}
\]

### Step 3: Write the Equation of the Tangent Line

With the slope \( m = -\frac{1}{2\sqrt{3}} \) and point \( \left( 2, -\frac{\pi}{6} \right) \), we use the point-slope form:
\[
y - y_1 = m(x - x_1)
\]
Substitute \( y_1 = -\frac{\pi}{6} \), \( x_1 = 2 \), and \( m = -\frac{1}{2\sqrt{3}} \):
\[
y + \frac{\pi}{6} = -\frac{1}{2\sqrt{3}}(x - 2)
\]

Simplifying, the equation of the tangent line is:
\[
y = -\frac{1}{2\sqrt{3}}x + \frac{1}{\sqrt{3}} - \frac{\pi}{6}
\]

Would you like further clarification or details on any step?

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**Related Questions:**

1. What is the general formula for the derivative of the inverse sine function?
2. How do you apply the chain rule in differentiation for inverse trigonometric functions?
3. What is the purpose of the point-slope form in finding the equation of a tangent?
4. How do you simplify expressions involving square roots in a denominator?
5. What does the graph of \( y = \sin^{-1} \left( -\frac{x}{4} \right) \) look like?

**Tip:** Always keep track of signs when differentiating, especially with trigonometric and inverse trigonometric functions, as they can change the slope direction.

Find the equation for the tangent to the graph of y at (2, -π/6) where y = sin⁻¹(-x/4).

Learn how to find the equation of the tangent line to the graph of y = sin⁻¹(-x/4) at the point (2, -π/6) with a step-by-step solution. This calculus problem covers differentiation, the chain rule, and inverse trigonometric functions. Suitable for students in Grades 11-12.

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