Question

50 points each question. Please help. How do I solve?

Answer 1

Here’s the Explanation in the image

Answer 2

`~~~~y= \sec^(-1) (14x)\\\\\\\implies (dy)/(dx) = (1)/(14x √((14x)^2 -1)) \cdot 14\\\\\\\implies (dy)/(dx)=(1)/(x√(196x^2 -1))\\\\\\\text{At point,}~~ (x_1,y_1) = \left((\sqrt2)/(14), \frac{\pi}4} \right), \text{slope of the tangent line,}\\\\\\m = (dy)/(dx) = \frac{1}{(\sqrt 2)/(14) \sqrt{196 \left((\sqrt 2)/(14) \right)^2-1}} \\\\\\~~~~~~~~~~~=(14)/(\sqrt 2 )\\\\\\~~~~~~~~~~~=(7\sqrt 2 \sqrt 2)/(\sqrt 2)\\\\\\`

`=7\sqrt 2\\\\\\\text{Equation of tangent line,}\\\\\\~~~~~y-y_1 = m(x -x_1)\\\\\\\implies y- \frac{\pi}4 = 7\sqrt 2 \left(x - (\sqrt 2)/(14) \right)\\ \\\\\implies y= 7x\sqrt 2 - 1+ \frac{\pi}4`