Answer
a) The tangent to the curve at $x=1$ is $y=\pi x-\pi+2$.
b) The smallest value of the slope of the curve is $\pi/2$.
Work Step by Step
$$y=f(x)=2\tan\Big(\frac{\pi x}{4}\Big)$$
a) Find $f'(x)$:
Using the Chain Rule: $$f'(x)=2\sec^2\Big(\frac{\pi x}{4}\Big)\Big(\frac{\pi x}{4}\Big)'=2\sec^2\Big(\frac{\pi x}{4}\Big)\Big(\frac{\pi }{4}\Big)=\frac{\pi}{2}\sec^2\Big(\frac{\pi x}{4}\Big)$$
For $x=1$, we have $$f(1)=2\tan\Big(\frac{\pi }{4}\Big)=2\times1=2$$
$$f'(1)=\frac{\pi}{2}\sec^2\Big(\frac{\pi }{4}\Big)=\frac{\pi}{2}\frac{1}{\cos^2\Big(\frac{\pi}{4}\Big)}=\frac{\pi}{2}\frac{1}{\Big(\frac{\sqrt2}{2}\Big)^2}=\frac{\pi}{2}\frac{1}{\frac{1}{2}}$$ $$=\frac{\pi}{2}\times2=\pi$$
As $f'(1)$ is the slope of the tangent line to $f(x)$ at $x=1$, we can find the tangent to $f(x)$ at $x=1$ following the formula:
$$y-f(1)=f'(1)(x-1)$$ $$y-2=\pi(x-1)$$ $$y=\pi x-\pi+2$$
b) We have $$f'(x)=\frac{\pi}{2}\sec^2\Big(\frac{\pi x}{4}\Big)=\frac{\pi}{2}\frac{1}{\cos^2\Big(\frac{\pi x}{4}\Big)}$$
The smallest value the slope can achieve happens when $f'(x)$ also reaches its smallest value, as $f'(x)$ represents the slope of the curve.
We see that $$\cos^2\Big(\frac{\pi x}{4}\Big)\le1$$ $$\frac{\pi}{2}\frac{1}{\cos^2\Big(\frac{\pi x}{4}\Big)}\ge\frac{\pi}{2}$$ $$f'(x)\ge\frac{\pi}{2}$$
So the minimum value of $f'(x)$ is $\pi/2$, which is also the smallest value of the slope of the curve.
