Answer
$$y''=2\sec^2\frac{x}{3}\tan\frac{x}{3}$$
Work Step by Step
$$y=9\tan\Big(\frac{x}{3}\Big)$$
- Find $y'$: $$y'=9\Bigg[\tan\Big(\frac{x}{3}\Big)\Bigg]'=9\sec^2\Big(\frac{x}{3}\Big)\Big(\frac{x}{3}\Big)'=9\sec^2\Big(\frac{x}{3}\Big)\times\frac{1}{3}$$ $$y'=3\sec^2\Big(\frac{x}{3}\Big)$$
- Find $y''$: $$y''=3\Bigg[\sec^2\Big(\frac{x}{3}\Big)\Bigg]'=6\sec\Big(\frac{x}{3}\Big)\Big(\sec\frac{x}{3}\Big)'=6\sec\frac{x}{3}\sec\frac{x}{3}\tan\frac{x}{3}\Big(\frac{x}{3}\Big)'$$ $$y''=6\sec^2\frac{x}{3}\tan\frac{x}{3}\times\frac{1}{3}$$ $$y''=2\sec^2\frac{x}{3}\tan\frac{x}{3}$$
