find the equation for the tangent to the graph of y at $(3,\\frac{\\pi}{6})$ \n$y = \\sin^{-1}(\\frac{x}{6})$…

find the equation for the tangent to the graph of y at $(3,\\frac{\\pi}{6})$ \n$y = \\sin^{-1}(\\frac{x}{6})$ \nthe equation of the line tangent to the graph of y at $(3,\\frac{\\pi}{6})$ is $y = \\square$ \n(uae integers or fractions for any numbers in the expression. type an exact answer, using radicals as needed.)

find the equation for the tangent to the graph of y at $(3,\\frac{\\pi}{6})$ \n$y = \\sin^{-1}(\\frac{x}{6})$ \nthe equation of the line tangent to the graph of y at $(3,\\frac{\\pi}{6})$ is $y = \\square$ \n(uae integers or fractions for any numbers in the expression. type an exact answer, using radicals as needed.)

Answer

Explanation:

Step1: Find the derivative of (y = \sin^{-1}\left(\frac{x}{6}\right))

Using the formula (\frac{d}{dx}\sin^{-1}(u)=\frac{1}{\sqrt{1 - u^{2}}}\cdot\frac{du}{dx}), where (u=\frac{x}{6}) and (\frac{du}{dx}=\frac{1}{6}). So (y^\prime=\frac{1}{\sqrt{1-\left(\frac{x}{6}\right)^{2}}}\cdot\frac{1}{6}=\frac{1}{\sqrt{36 - x^{2}}})

Step2: Evaluate the derivative at (x = 3)

Substitute (x = 3) into (y^\prime). (y^\prime(3)=\frac{1}{\sqrt{36-9}}=\frac{1}{\sqrt{27}}=\frac{1}{3\sqrt{3}}=\frac{\sqrt{3}}{9})

Step3: Use the point - slope form (y - y_{0}=m(x - x_{0}))

Here (x_{0}=3,y_{0}=\frac{\pi}{6},m = \frac{\sqrt{3}}{9}) (y-\frac{\pi}{6}=\frac{\sqrt{3}}{9}(x - 3)) (y=\frac{\sqrt{3}}{9}x-\frac{\sqrt{3}}{3}+\frac{\pi}{6})

Answer:

(y=\frac{\sqrt{3}}{9}x-\frac{\sqrt{3}}{3}+\frac{\pi}{6})