Calculus: Early Transcendentals (2nd Edition)

The equation of the tangent line is $y=-2\sqrt{3}x+\dfrac{2\pi\sqrt{3}+3}{3}$
$y=\dfrac{\cos x}{1-\cos x}$ $;$ $x=\dfrac{\pi}{3}$ Evaluate the derivative of the function given using the quotient rule: $y'=\dfrac{(1-\cos x)(\cos x)'-\cos x(1-\cos x)'}{(1-\cos x)^{2}}=...$ Evaluate the derivatives indicated in the numerator and simplify: $...=\dfrac{-(1-\cos x)(\sin x)-\cos x(\sin x)}{(1-\cos x)^{2}}=...$ $...=\dfrac{-\sin x+\sin x\cos x-\sin x\cos x}{(1-\cos x)^{2}}=...$ $...=-\dfrac{\sin x}{(1-\cos x)^{2}}$ Substitute $x=\dfrac{\pi}{3}$ in the derivative found and simplify to obtain the slope of the tangent line at the given point: $m=-\dfrac{\sin\dfrac{\pi}{3}}{\Big(1-\cos\dfrac{\pi}{3}\Big)^{2}}=-\dfrac{\dfrac{\sqrt{3}}{2}}{\Big(1-\dfrac{1}{2}\Big)^{2}}=-\dfrac{\dfrac{\sqrt{3}}{2}}{\dfrac{1}{4}}=-\dfrac{4\sqrt{3}}{2}=-2\sqrt{3}$ Substitute $x=\dfrac{\pi}{3}$ in the original expression to obtain the $y$-coordinate of the point given: $y$-coordinate $=\dfrac{\cos\dfrac{\pi}{3}}{1-\cos\dfrac{\pi}{3}}=\dfrac{\dfrac{1}{2}}{1-\dfrac{1}{2}}=\dfrac{\dfrac{1}{2}}{\dfrac{1}{2}}=1$ The point is $\Big(\dfrac{\pi}{3},1\Big)$ 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: $y-1=-2\sqrt{3}\Big(x-\dfrac{\pi}{3}\Big)$ $y-1=-2\sqrt{3}x+\dfrac{2\pi\sqrt{3}}{3}$ $y=-2\sqrt{3}x+\dfrac{2\pi\sqrt{3}}{3}+1$ $y=-2\sqrt{3}x+\dfrac{2\pi\sqrt{3}+3}{3}$ The graph of the function and the tangent line is shown in the answer section.