I guess we are doing dt/du?
remember the power rule and stuff
[tex]\frac{dy}{dx} x^m=mx^{m-1}[/tex]
oh, also [tex]\sqrt[n]{x^m}=x^\frac{m}{n}[/tex]
so
[tex]u=\sqrt[3]{t^2}-6\sqrt[2]{t^3}[/tex]
[tex]u=\sqrt[3]{t^2}-6t\sqrt[2]{t}[/tex]
[tex]u=t^\frac{2}{3}-6t(t)^\frac{1}{2}[/tex]
differentiate
[tex]\frac{dt}{du}=\frac{2t^\frac{-1}{3}}{3}-\frac{6t(t)^\frac{-1}{2}}{2}[/tex]
[tex]\frac{dt}{du}=\frac{2}{3t^\frac{1}{3}}-\frac{3t}{t^\frac{1}{2}}[/tex]
[tex]\frac{dt}{du}=\frac{2}{3\sqrt[3]{t}}-\frac{3t}{\sqrt{t}}[/tex]
[tex]\frac{dt}{du}=\frac{2\sqrt[3]{t^2}}{3\sqrt[3]{t^3}}-\frac{3t\sqrt{t}}{\sqrt{t^2}}[/tex]
[tex]\frac{dt}{du}=\frac{2\sqrt[3]{t^2}}{3t}-\frac{3t\sqrt{t}}{t}[/tex]
[tex]\frac{dt}{du}=\frac{2\sqrt[3]{t^2}}{3t}-\frac{9t\sqrt{t}}{3t}[/tex]
[tex]\frac{dt}{du}=\frac{2\sqrt[3]{t^2}-9t\sqrt{t}}{3t}[/tex]
