Answer:
[tex]\frac{dy}{dx}=\frac{x(1-2lnx)}{x^{4}}[/tex]
Step-by-step explanation:
To solve the question we refresh our knowledge of the quotient rule.
For a function f(x) express as a ratio of another functions u(x) and v(x) i.e
[tex]f(x)=\frac{u(x)}{v(x)}\\[/tex], the derivative is express as
[tex]\frac{df(x)}{dx}=\frac{v(x)\frac{du(x)}{dx}-u(x)\frac{dv(x)}{dx}}{v(x)^{2} }[/tex]
from [tex]y=lnx/x^{2}[/tex]
we assign u(x)=lnx and v(x)=x^2
and the derivatives
[tex]\frac{du(x)}{dx}=\frac{1}{x}\\\frac{dv(x)}{dx}=2x\\[/tex].
Note the expression used in determining the derivative of the logarithm function.it was obtain from the general expression of logarithm derivative i.e [tex]y=lnx\\\frac{dy}{dx}=\frac{1}{x}[/tex]
If we substitute values into the quotient expression we arrive at
[tex]\frac{dy}{dx}=\frac{(x^{2}*\frac{1}{x})-(2x*lnx)}{x^{4}}\\\frac{dy}{dx}=\frac{x-2xlnx}{x^{4}}\\\frac{dy}{dx}=\frac{x(1-2lnx)}{x^{4}}[/tex]
