Find an equation of the tangent line to the curve at the given point. y=sin(sinx) , (2pi,0)

Find an equation of the tangent line to the curve at the given point.
$y=\mathrm{sin}\left(\mathrm{sin}x\right),\left(2\pi ,0\right)$
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Cristiano Sears
The derivative ${y}^{\prime }=\mathrm{cos}x\mathrm{cos}\left(\mathrm{sin}x\right)$ by the product rule.
At $x=2\left(\pi \right),{y}^{\prime }=\mathrm{cos}\left(0\right)\cdot \mathrm{cos}\left(0\right)=1,\because \mathrm{sin}\left(2\left(\pi \right)\right)=0.$
Therefore 1 is the slope of the tangent and the slope of the tangent line y=x+b, where b is the y intercept.
Putting in the point $\left(2\left(\pi \right),0\right):0=2\left(\pi \right)+b,sob=-2\left(\pi \right)\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}y=x-2\left(\pi \right).$