Question

find the equation of the tangent line to the curve ( y = 2\tan x ) at the point ( (pi/4,2) ). the equation of this tangent line can be written in the form ( y = mx + b ) where ( m ) is: and where ( b ) is:

Explanation:

Step1: Find the derivative of the function

The derivative of \(y = 2\tan x\) is \(y'=2\sec^{2}x\).

Step2: Evaluate the derivative at \(x = \frac{\pi}{4}\)

When \(x=\frac{\pi}{4}\), \(\sec x=\sqrt{2}\), so \(y'|_{x = \frac{\pi}{4}}=2\times(\sqrt{2})^{2}=4\). Thus, \(m = 4\).

Step3: Use the point - slope form to find \(b\)

The equation of the line is \(y - y_{0}=m(x - x_{0})\), substituting \(x_{0}=\frac{\pi}{4}\), \(y_{0} = 2\) and \(m = 4\): \(y-2=4(x-\frac{\pi}{4})\).
Simplify to \(y=4x + 2-\pi\). So \(b=2-\pi\).

Answer:

\(m = 4\), \(b=2-\pi\)