# `y = arcsec(4x) , (sqrt(2)/4, pi/4)` Find an equation of the tangent line to the graph of the function at the given point

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Equation of a tangent line to the graph of function `f` at point `(x_0,y_0)` is given by `y=y_0+f'(x_0)(x-x_0).`

The first step to finding equation of tangent line is to calculate the derivative of the given function. To calculate this derivative we will have to use the *chain rule* `(u(v))'=u'(v)cdot v'.`

`y'=1/(|4x|sqrt((4x)^2-1))cdot4=1/(|x|sqrt(16x^2-1))`

Now we calculate the value of the derivative at the given point.

`y'(sqrt2/4)=1/(|sqrt2/4|sqrt(16(sqrt2/4)^2-1))=1/(sqrt2/4sqrt(16cdot1/8-1))=1/(sqrt2/4)=4/sqrt2=2sqrt2`

We now have everything needed to write the equation of the tangent line.

`y=pi/4+2sqrt2(x-sqrt2/4)`

`y=2sqrt2x+(pi-4)/4`

Graph of the function (red) along with the tangent line (blue) can be seen in the image below.