`y = 2(tan(x))^3, ((pi/4),2)` Find an equation of the tangent line to the graph of f at the given point.

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Chapter 2, 2.4 - Problem 80 - Calculus of a Single Variable (10th Edition, Ron Larson).
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You need to find the equation of the tangent line to the given curve, at the point (pi/4,2), using the formula:

`f(x) - f(pi/4) = f'(pi/4)(x - pi/4)`

You need to notice that `f(pi/4) = 2.`

You need to evaluate the derivative of the given function, using chain rule, such that:

`f'(x) = (2(tan x)^3)' => f'(x) = 2*3(tan x)^2*(tan x)'`

`f'(x) = 6tan^2 x*(1/(cos^2 x))`

You need to evaluate f'(x) at `x = pi/4` , hence, you need to replace `pi/4` for x in equation of derivative:

`f'(pi/4) = 6tan^2 (pi/4)*(1/(cos^2 (pi/4)))`

`f'(pi/4) = 6*1/(1/2) => f'(pi/4) = 12`

You need to replace the values into equation of tangent line, such that:

`f(x) - 2 = 12(x - pi/4) => f(x) = 12x - 3pi + 2 `

Hence, evaluating the equation of the tangent line to the given curve, at the given point, yields `f(x) = 12x - 3pi + 2` .

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