Find the linear approximation to y=arcsinh(2x) at x=6.

            $$\begin{gathered} \mathrm{Given}: \\ \mathrm{y}=\mathrm{arcsinh}\left(2\mathrm{x}\right) \mathrm{and} {\mathrm{x}}_0=6 \\ \mathrm{To} \mathrm{find}: \\ \mathrm{The} \mathrm{linear} \mathrm{approximation} \mathrm{to} \mathrm{y}=\mathrm{arcsinh}\left(2\mathrm{x}\right) \mathrm{at} {\mathrm{x}}_0=6 \end{gathered}$$

           $$\begin{gathered} A linear approximation is given by \\ L\left(x\right)\approx f\left(x_0\right)+f\text{'}\left(x_0\right)(x-x_0) ...\left(1\right) \\ Given that y=f\left(x\right)=arc\sin h\left(2x\right) and x_0=6 \\ Now, \\ {y}_0=f\left(x_0\right)=arc\sin h\left(2x_0\right) \\ {y}_0=f\left(x_0\right)=arc\sin h\left[\left(2\right)\left(6\right)\right] \\ f\left(x_0\right)=arc\sin h\left(12\right) \\ find the deivative of function f\left(x\right)=arc\sin h\left(2x\right) \\ f\text{'}\left(x\right)=\frac{2}{\sqrt{4x^2+1}} \\ f\text{'}\left(x_0\right)=\frac{2}{\sqrt{4{x_0}^2+1}} \\ f\text{'}\left(x_0\right)=\frac{2}{\sqrt{145}} ...(x_0=6) \end{gathered}$$