## University Calculus: Early Transcendentals (3rd Edition)

${dy}={4{x^2}}\sec^2({\frac{{x}^3}{3}}){dx}$
We evaluate the function: $y=4\tan(\frac{{x}^3}{3})$ on differentiating the above: $\frac{dy}{dx}=\frac{d(4\tan(\frac{{x}^3}{3}))}{dx}$ or $\frac{dy}{dx}=4\sec^2(\frac{{x}^3}{3})\frac{d\frac{{x}^3}{3}}{dx}$ or $\frac{dy}{dx}=4\frac{3x^2}{3}\sec^2(\frac{{x}^3}{3})$ or ${dy}={4{x^2}}\sec^2({\frac{{x}^3}{3}}){dx}$ The final answer is: ${dy}={4{x^2}}\sec^2({\frac{{x}^3}{3}}){dx}$