Answer :
Given:
The function is,
[tex]g(x)=\csc ^{-1}(7x^2)[/tex]First find the derivative,
[tex]\begin{gathered} g^{\prime}(x)=\frac{d}{dx}(\csc ^{-1}(7x^2)) \\ Use\text{ the derivative,} \\ \frac{d}{dx}(\csc ^{-1}(x))=\frac{-1}{x\sqrt[]{x^2-1}} \\ g^{\prime}(x)=-\frac{1}{(7x^2)\sqrt[]{(7x^2)-1}}\times\frac{d}{dx}(7x^2) \\ g^{\prime}(x)=-\frac{14x}{7x^2\sqrt[]{49x^4-1}} \\ g^{\prime}(x)=-\frac{2}{x\sqrt{49x^4-1}} \end{gathered}[/tex]Now put the given point in the derivative to obtain the slope,
[tex]\begin{gathered} g^{\prime}(x)=-\frac{2}{x\sqrt[]{49x^4-1}}atx=\frac{\sqrt[]{2}}{\sqrt[]{7}}^{} \\ slope=-\frac{2}{\frac{\sqrt[]{2}}{\sqrt[]{7}}\sqrt[]{49(\frac{\sqrt[]{2}}{\sqrt[]{7}})^4-1}} \\ =-\frac{2}{\sqrt{\frac{6}{7}}} \\ =-\frac{2\sqrt{7}}{\sqrt{6}} \\ =-\frac{\sqrt{2}\sqrt{7}}{\sqrt{3}} \\ =-\sqrt{\frac{14}{3}} \end{gathered}[/tex]Now to find the y coordinate
[tex]\begin{gathered} \text{Put x=}\frac{\sqrt[]{2}}{\sqrt[]{7}}\text{ in the given function} \\ g(x)=y \\ y=\csc ^{-1}(7x^2) \\ y=\csc ^{-1}(7(\frac{\sqrt[]{2}}{\sqrt[]{7}}^{})^2) \\ y=\csc ^{-1}(2) \\ y=0.5236 \end{gathered}[/tex]The point is,
[tex](x,y)=(2,0.5236)[/tex]Use point-slope form,
[tex]\begin{gathered} y-y_0=m(x-x_0) \\ y-0.5236=-\sqrt[]{\frac{14}{3}}(x-2) \\ y=-\sqrt[]{\frac{14}{3}}x+2\sqrt[]{\frac{14}{3}}+0.5236 \\ y=-\sqrt[]{\frac{14}{3}}x+4.8441 \end{gathered}[/tex]Answer: The equation of tangent line is,
[tex]y=-\sqrt[]{\frac{14}{3}}x+4.8441[/tex]