## Calculus: Early Transcendentals (2nd Edition)

The equation of the tangent line is $y=\sqrt{3}x+\dfrac{12-\pi\sqrt{3}}{6}$
$y=1+2\sin x$ $;$ $x=\dfrac{\pi}{6}$ Evaluate the derivative of the function given: $y'=2\cos x$ Substitute $x=\dfrac{\pi}{6}$ in the derivative found and simplify to obtain the slope of the tangent line at the given point: $m=2\cos\dfrac{\pi}{6}=2\Big(\dfrac{\sqrt{3}}{2}\Big)=\sqrt{3}$ Substitute $x=\dfrac{\pi}{6}$ in the original expression to obtain the $y$-coordinate of the point given: $y$-coordinate $=1+2\sin\dfrac{\pi}{6}=1+2\Big(\dfrac{1}{2}\Big)=1+1=2$ The point is $\Big(\dfrac{\pi}{6},2\Big)$ The slope of the tangent line and a point through which it passes are now known. Use the point-slope form of the equation of a line, which is $y-y_{1}=m(x-x_{1})$ to obtain the equation of the tangent line at the given point: $y-2=\sqrt{3}\Big(x-\dfrac{\pi}{6}\Big)$ $y-2=\sqrt{3}x-\dfrac{\pi\sqrt{3}}{6}$ $y=\sqrt{3}x-\dfrac{\pi\sqrt{3}}{6}+2$ $y=\sqrt{3}x+\dfrac{12-\pi\sqrt{3}}{6}$ The graph of the function and the tangent line is shown in the answer section.