find the equation of the tangent line to the curve y = 6 sec x - 12 cos x at the point (π/3, 6). find the…

Explanation:

Step1: Differentiate the function

We know that $\frac{d}{dx}(\sec x)=\sec x\tan x$ and $\frac{d}{dx}(\cos x)=-\sin x$. So if $y = 6\sec x-12\cos x$, then $y'=6\sec x\tan x + 12\sin x$.

Step2: Find the slope of the tangent - line

Substitute $x = \frac{\pi}{3}$ into $y'$. We know that $\sec\frac{\pi}{3}=2$, $\tan\frac{\pi}{3}=\sqrt{3}$, and $\sin\frac{\pi}{3}=\frac{\sqrt{3}}{2}$. Then $y'\left(\frac{\pi}{3}\right)=6\times2\times\sqrt{3}+12\times\frac{\sqrt{3}}{2}=12\sqrt{3}+6\sqrt{3}=18\sqrt{3}$. So the slope of the tangent - line is $m = 18\sqrt{3}$.

Step3: Use the point - slope form to find the equation of the tangent line

The point - slope form of a line is $y - y_1=m(x - x_1)$, where $(x_1,y_1)=\left(\frac{\pi}{3},6\right)$ and $m = 18\sqrt{3}$. So $y-6 = 18\sqrt{3}\left(x-\frac{\pi}{3}\right)$, which simplifies to $y=18\sqrt{3}x-6\pi\sqrt{3}+6$.

Answer:

The equation of this tangent line has a slope of: $18\sqrt{3}$ The equation of the tangent line is: $y = 18\sqrt{3}x-6\pi\sqrt{3}+6$