![\bf y=\cfrac{5}{2\sqrt[4]{x^3}}\implies y=\cfrac{5}{2}\cdot x^{-\frac{3}{4}}\\\\ -----------------------------\\\\ \cfrac{dy}{dx}=\cfrac{5}{2}\left( -\cfrac{3}{4}x^{-\frac{3}{4}-1} \right)\implies \cfrac{dy}{dx}=\cfrac{5}{2}\left( -\cfrac{3}{4}x^{-\frac{7}{4}} \right) \\\\\\ \cfrac{dy}{dx}=-\cfrac{15}{8}\cdot x^{-\frac{7}{4}} \implies \cfrac{dy}{dx}=-\cfrac{15}{8}\cdot \cfrac{1}{x^{\frac{7}{4}} } \\\\\\ ](/tpl/images/0467/6456/55e50.png)
![\bf \cfrac{dy}{dx}=-\cfrac{15}{8x^{\frac{7}{4}}}\implies \cfrac{dy}{dx}=-\cfrac{15}{8\sqrt[4]{x^7}}\\\\\\ \cfrac{dy}{dx}=-\cfrac{15}{8x\sqrt[4]{x^3}} ](/tpl/images/0467/6456/b8c3a.png)
Find the derivative.
y = 5/ (2root4(x^3))...
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