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 $6 \pi x +12y =6\sqrt{3}+ \pi$ 

Start with your answer: $y = -\dfrac{\pi x}{2} + \dfrac{\sqrt{3}}{2} + \dfrac{\pi}{12}$
Multiply both sides by $12$: $12y = -6\pi x + 6\sqrt{3} + \pi$
Add $6 \pi x$ to both sides: $6\pi x + 12y = 6\sqrt{3} + \pi$
Therefore your answer is equivalant to the book's answer. 
