Answer and Step-by-step explanation:
[tex]Greetings![/tex]
[tex]Let's~answer~your~question![/tex]
→ [tex]By~quotient~rule,[/tex]
[tex]y'=\frac{\frac{1}{x}*x~-~In~x*1 }{x^2} =\frac{1-In~x}{x^2}[/tex]
→ [tex]This~ problem ~can ~also ~be ~solved ~by~ the ~Product~ Rule:[/tex]
[tex]y'=f'(x)~g(x)+f(x)~g(x)[/tex]
→ [tex]The~ original~ function ~can~ also~be~ rewritten~ using ~negative~ exponents:[/tex]
[tex]f(x)=\frac{In(x)}{x}=In(x)*x^-^1[/tex]
[tex]f'(x)=\frac{1}{x} *x^-^1+ln(x)*-1x^-^2[/tex]
[tex]f'(x)=\frac{1}{x}*\frac{1}{x}+In(x)*-\frac{1}{x^2}[/tex]
[tex]f'(x)=\frac{1}{x^2}-\frac{In(x)}{x^2}[/tex]
[tex]f'(x)=\frac{1-In(x)}{x^2}[/tex]
