Question

Find an equation of the line tangent to g(x)=csc^-1(7x^2) at x=root2/root7. Use point slope form.

Answer 1

Given:

The function is,

`g(x)=\csc ^(-1)(7x^2)`

First find the derivative,

`\begin{gathered} g^(\prime)(x)=(d)/(dx)(\csc ^(-1)(7x^2)) \\ Use\text{ the derivative,} \\ (d)/(dx)(\csc ^(-1)(x))=\frac{-1}{x\sqrt[]{x^2-1}} \\ g^(\prime)(x)=-\frac{1}{(7x^2)\sqrt[]{(7x^2)-1}}*(d)/(dx)(7x^2) \\ g^(\prime)(x)=-\frac{14x}{7x^2\sqrt[]{49x^4-1}} \\ g^(\prime)(x)=-(2)/(x√(49x^4-1)) \end{gathered}`

Now put the given point in the derivative to obtain the slope,

`\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{(6)/(7)}} \\ =-(2√(7))/(√(6)) \\ =-(√(2)√(7))/(√(3)) \\ =-\sqrt{(14)/(3)} \end{gathered}`

Now to find the y coordinate

`\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}`

The point is,

`(x,y)=(2,0.5236)`

Use point-slope form,

`\begin{gathered} y-y_0=m(x-x_0) \\ y-0.5236=-\sqrt[]{(14)/(3)}(x-2) \\ y=-\sqrt[]{(14)/(3)}x+2\sqrt[]{(14)/(3)}+0.5236 \\ y=-\sqrt[]{(14)/(3)}x+4.8441 \end{gathered}`

Answer: The equation of tangent line is,

`y=-\sqrt[]{(14)/(3)}x+4.8441`