Answer:
[tex]Y-\frac{\pi}{2} =\frac{\pi}{2} (x+\sqrt{\frac{1}{7}})[/tex]
Step-by-step explanation:
The equation of the curve is
[tex]Y = sin^{-1}(7x)[/tex]
To find the equation of tangent we need to differentiate this equation w.r.t x
So, differentiating we get
[tex]Y'=\frac{7}{\sqrt{1-49x^2} }[/tex]
This would give the slope of the tangent line at any given point of which x coordinate is known. In the present case it is [tex]x = \sqrt{\frac{1}{7} }[/tex]
Then slope would accordingly be
[tex]Y'=\frac{7}{\sqrt{1-49/49} }[/tex]
= ∞
For, [tex]x = \sqrt{\frac{1}{7} }[/tex], [tex]Y = sin^{-1}(7/7)= \pi/2[/tex]
Equation of tangent line, in the point slope form, would be [tex]Y-\frac{\pi}{2} =\frac{\pi}{2} (x+\sqrt{\frac{1}{7}})[/tex]
