a) [tex]\frac{dy}{dx} = \frac{1-3lnx}{x^4}[/tex]
b) [tex]\frac{dy}{dx} = \frac{-2}{3\sqrt[3]{1-x} }[/tex]
Explanation:
a) [tex]y=\frac{lnx}{x^3}[/tex]
[tex]y = \frac{lnx}{\sqrt[3]{x} }[/tex]
[tex]y = lnx. x^-3[/tex]
Differentiating the above equation in terms of x
[tex]\frac{dy}{dx} = \frac{1}{x} \times x^-3 - 3lnx\times x^-^4\\\frac{dy}{dx} = \frac{1}{x^4} - \frac{3lnx}{x^4} \\\frac{dy}{dx} = \frac{1-3lnx}{x^4} \\[/tex]
b) [tex]y = \sqrt[3]{1-x^{2} }[/tex]
Differentiating the above equation in terms of x
[tex]y = \sqrt[3]{(1-x)^{2} } \\\frac{dy}{dx} = (1-x)^\frac{2}{3} \\\frac{dy}{dx} = \frac{2}{3}\times (1-x)^\frac{-1}{3} \times -1\\\frac{dy}{dx} = \frac{-2}{3\sqrt[3]{1-x} }[/tex]
Thus,
a) [tex]\frac{dy}{dx} = \frac{1-3lnx}{x^4}[/tex]
b) [tex]\frac{dy}{dx} = \frac{-2}{3\sqrt[3]{1-x} }[/tex]
