derivatives
Given curve $y=2\tan(\pi x/4)$, find tangent line equation when $x=1$
$$\frac{dy}{dx}= 2\frac \pi 4 \sec^2\left(\frac{\pi x} 4 \right) = \frac{\pi2}{4\cdot2} =\pi$$ when $x=1$
so how do I find the tangent line equation,
I only know that $y=mx+c$, thus $y=\pi x+c$?
If we suppose that your curve is the graph of a function $y=f(x)$ such that $f(0) = \sqrt{\pi/2}$, than the equation of the tangent at $x=0$ is:
$ y-\sqrt{\pi/2}=f'(0)(x-0) $
i.e.
$y=f'(0) x+\sqrt{\pi/2}$
The slope of the tangent line is $$m=y'=\frac{-2}{(2x-1)^2}$$ and since it is perpendicular to the line $x-2y-1=0$ whose slope is $1/2$ we get $m=-2$ and hence, $$-2=\frac{-2}{(2x-1)^2}$$ then solve for $x$. You must get $x=0$ or $x=1$. The points of tangency are $(0,-1)$ and $(1,1)$. The two tangent lines must be $2x+y+1=0$ and $2x+y-3=0$. See the graph below.
Best Answer
Find $y$ at $ x = 1 ;$ ...as $ y = 2 $
The point of tangency is
$$ (1, 2) $$
So the tangent equation is
$$ \frac{y- 2}{x-1} = \pi. $$