## Calculus: Early Transcendentals (2nd Edition)

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