<s>[INST] Find the tangen to $\cos(\pi \cdot x)$ I have the following assignment.
Find the tangent to $y=f(x)=\cos(\pi \cdot x)$ at $x=\displaystyle\frac{1}{6}$.
First step would be to take the derivative of $f(x)$
$f'(x)= -\pi \sin(\pi \cdot x)$
Then I put the $x$-value into $f'(x)$ to find the slope
$f'(\displaystyle\frac{1}{6})= -\pi \sin(\pi \cdot \frac{1}{6})= \frac{-\pi}{2}$
And I put the $x$-value into $f(x)$ to find the $y$-value
$f(\displaystyle\frac{1}{6})=\cos(\pi \cdot \frac{1}{6})=\frac{\sqrt{3}}{2}$
Now I use the formula $y=m(x-x_0)+y_0$
$y= \displaystyle\frac{-\pi}{2}(x-\frac{1}{6})+\frac{\sqrt{3}}{2}$ = $\displaystyle\frac{-\pi \cdot x}{2} + \frac{\sqrt{3}}{2} + \frac{\pi}{12}$
Am I correct? Because my book got an different answer like $6 \pi x +12y =6\sqrt{3}+ \pi$ 
 [/INST] multiplying your equation
$y=-\frac{\pi}{2}x+\frac{\sqrt{3}}{2}+\frac{\pi}{12}$ by $12$ you will get
$12y+6\pi x=6\sqrt{3}+\pi$
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