1. ## differentiate

differentiate

Differentiate.

2. $\displaystyle g(t) = \frac{8t-\sqrt{t}}{t^\frac{1}{3}}$

$\displaystyle u = 8t - \sqrt{t}$

$\displaystyle v = t^\frac{1}{3}$

$\displaystyle u' = 8 - \frac{1}{2}t^\frac{-1}{2}$

$\displaystyle v' = \frac{1}{3}t^\frac{-2}{3}$

$\displaystyle = \frac{t^\frac{1}{3}.(8- \frac{1}{2}t^\frac{-1}{2}) - (8t - \sqrt{t}).(\frac{1}{3}t^\frac{-2}{3})}{\frac{1}{4}t^\frac{-4}{3}}$

$\displaystyle = \frac{8t^\frac{1}{3} - \frac{1}{2}t^\frac{-1}{6} - \frac{8}{3}t^\frac{1}{3} + \frac{1}{3}t^\frac{-1}{6}}{\frac{1}{4}t^\frac{-4}{3}}$

$\displaystyle = \frac{\frac{16}{3}t^\frac{1}{3} - \frac{1}{6}t^\frac{-1}{6}}{\frac{1}{4}t^\frac{-4}{3}}$

3. OH MY...

$\displaystyle v = (5\sqrt{x}+ \frac{6}{5\sqrt[3]{x}})^2$

$\displaystyle u = 5\sqrt{x}+ \frac{6}{5\sqrt[3]{x}}$

$\displaystyle v = u^2$

$\displaystyle \frac{dv}{du} = 2u$

$\displaystyle u = 5x^\frac{1}{2} + \frac{6}{5}x^\frac{-1}{3}$

$\displaystyle \frac{du}{dx} = \frac{5}{2}x^\frac{-1}{2} -\frac{-6}{15}x^\frac{-4}{3}$

$\displaystyle \frac{dv}{dx} = \frac{du}{dx}.\frac{dv}{du}$

$\displaystyle = 2u(\frac{5}{2}x^\frac{-1}{2} \frac{-6}{15}x^\frac{-4}{3})$

$\displaystyle = \frac{5u}{\sqrt{x}} - \frac{4u}{5\sqrt[3]{x^4}}$

$\displaystyle = \frac{5(5\sqrt{x}+ \frac{6}{5\sqrt[3]{x}}) }{\sqrt{x}} - \frac{4(5\sqrt{x}+ \frac{6}{5\sqrt[3]{x}}) }{5\sqrt[3]{x^4}}$

$\displaystyle \frac{25\sqrt{x} + \frac{30}{5\sqrt[3]{x}}}{\sqrt{x}} - \frac{20\sqrt{x} + \frac{24}{5\sqrt[3]{x}}}{5\sqrt[3]{x^4}}$

$\displaystyle = 25 + \frac{\frac{30}{5\sqrt[3]{5}}}{\sqrt{x}} - \frac{20\sqrt{x}}{5\sqrt[3]{x^4}} + 24$

$\displaystyle = 49 + \frac{\frac{30}{5\sqrt[3]{5}}}{\sqrt{x}} - \frac{20\sqrt{x}}{5\sqrt[3]{x^4}}$

$\displaystyle = 49 + \frac{\frac{6}{\sqrt[3]{5}}}{\sqrt{x}} - \frac{4\sqrt{x}}{\sqrt[3]{x^4}}$