Question

HELP!! PLS!! MEDALS! BASIC CALC QUESTION



Find an equation of the line tangent to the graph of f at the given point.

f(x)=arctan(2x); (1/2, π/4)

Answer 1

the answer is 2x-4y=2-pi

Answer 2

First find derivative f'(x)
rewrite equation as
\[\tan(y) = 2x\]
Differentiate
\[\sec^2 (y) \frac{dy}{dx} = 2\]
\[(1+\tan^2 (y)) \frac{dy}{dx} = 2\]
\[\frac{dy}{dx} = \frac{2}{1+ \tan^2 (\tan^{-1} 2x)} = \frac{2}{1+(2x)^2} = \frac{2}{1+4x^2}\]
The equation of a line is:
\[y = \frac{dy}{dx} (x-\frac{1}{2}) + \frac{\pi}{4}\]
plugging in x=1/2
dy/dx = 1
Therefore final answer is:
\[y = x-\frac{1}{2} + \frac{\pi}{4}\]