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marin [14]
3 years ago
9

Find the tangent line equation of the curve at the given point. Y=arcsin(7x) at the point where x=sqrt2/14

Mathematics
1 answer:
Mumz [18]3 years ago
4 0

Answer:

Y-\frac{\pi}{2} =\frac{\pi}{2} (x+\sqrt{\frac{1}{7}})

Step-by-step explanation:

The equation of the curve is

Y = sin^{-1}(7x)

To find the equation of tangent we need to differentiate this equation w.r.t x

So, differentiating we get

Y'=\frac{7}{\sqrt{1-49x^2} }

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  x = \sqrt{\frac{1}{7} }

Then slope would accordingly be

Y'=\frac{7}{\sqrt{1-49/49} }

= ∞

For, x = \sqrt{\frac{1}{7} }, Y = sin^{-1}(7/7)= \pi/2

Equation of tangent line, in the point slope form, would be Y-\frac{\pi}{2} =\frac{\pi}{2} (x+\sqrt{\frac{1}{7}})

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