Answer:
[tex] \orange{ \bold{\frac{dy}{dx} =\frac{ 5{x}^{2} + 3 }{3\sqrt[3]{(1 + {x}^{2})^{2} } }}}[/tex]
Step-by-step explanation:
[tex]y = x \sqrt[3]{1 + {x}^{2} } \\ assuming \: log \: both \: sides \\log y = log(x \sqrt[3]{1 + {x}^{2} } ) \\ \therefore log y = logx + log(\sqrt[3]{1 + {x}^{2} } ) \\ \therefore log y = logx + \frac{1}{3} log({1 + {x}^{2} } ) \\ differentiating \: both \: sides \: w.r.t.x \\ \frac{1}{y} \frac{dy}{dx} = \frac{1}{x} + \frac{1}{3} . \frac{1}{(1 + {x}^{2}) } (0 + 2x) \\ \frac{1}{y} \frac{dy}{dx} = \frac{1}{x} + \frac{2x}{3(1 + {x}^{2}) }\\ \frac{1}{y} \frac{dy}{dx} =\frac{3(1 + {x}^{2}) + 2 {x}^{2} }{3x(1 + {x}^{2}) }\\ \frac{1}{y} \frac{dy}{dx} =\frac{3 + 3{x}^{2} + 2 {x}^{2} }{3x(1 + {x}^{2}) }\\ \frac{1}{y} \frac{dy}{dx} =\frac{3 + 5{x}^{2} }{3x(1 + {x}^{2}) }\\ \frac{dy}{dx} =\frac{y(3 + 5{x}^{2} )}{3x(1 + {x}^{2}) } \\ \\ \frac{dy}{dx} =\frac{x \sqrt[3]{1 + {x}^{2} } (3 + 5{x}^{2} )}{3x(1 + {x}^{2}) }\\ \\ \frac{dy}{dx} =\frac{(3 + 5{x}^{2} )\sqrt[3]{1 + {x}^{2} } }{3(1 + {x}^{2}) }\\ \\ \purple{ \bold{\frac{dy}{dx} =\frac{ 5{x}^{2} + 3 }{3\sqrt[3]{(1 + {x}^{2})^{2} } }}}[/tex]
