Question

What is the derivative of ="f(x)=\frac{In(x)}{x}~?" alt="f(x)=\frac{In(x)}{x}~?" align="absmiddle" class="latex-formula">

Answer 1

Answer and Step-by-step explanation:

$Greetings!$

$Let's~answer~your~question!$

→ $By~quotient~rule,$

$y'=\frac{\frac{1}{x}*x~-~In~x*1 }{x^2} =\frac{1-In~x}{x^2}$

→ $This~ problem ~can ~also ~be ~solved ~by~ the ~Product~ Rule:$

$y'=f'(x)~g(x)+f(x)~g(x)$

→ $The~ original~ function ~can~ also~be~ rewritten~ using ~negative~ exponents:$

$f(x)=\frac{In(x)}{x}=In(x)*x^-^1$

$f'(x)=\frac{1}{x} *x^-^1+ln(x)*-1x^-^2$

$f'(x)=\frac{1}{x}*\frac{1}{x}+In(x)*-\frac{1}{x^2}$

$f'(x)=\frac{1}{x^2}-\frac{In(x)}{x^2}$

$f'(x)=\frac{1-In(x)}{x^2}$