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Derivative of ln(x)/x using the quotient rule

Let y=ln⁡(x)x.y = \frac{\ln(x)}{x}. Use the quotient rule, along with the fact that ddx[ln⁡x]=1x\frac{d}{dx}[\ln x] = \frac{1}{x}, to find dydx\dfrac{dy}{dx}. Simplify your answer as much as possible.

Hint

Write y=f(x)/g(x)y = f(x)/g(x) with f(x)=ln⁡xf(x) = \ln x and g(x)=xg(x) = x. Apply (fg)′=f′g−fg′g2\left(\frac{f}{g}\right)' = \frac{f'g - fg'}{g^2}, using f′(x)=1/xf'(x) = 1/x and g′(x)=1g'(x) = 1, then combine terms in the numerator over a common denominator before simplifying.

Step-by-step solution

Identify f(x)=ln⁡xf(x) = \ln x and g(x)=xg(x) = x, so f′(x)=1xf'(x) = \frac{1}{x} and g′(x)=1g'(x) = 1. The quotient rule gives dydx=f′(x)g(x)−f(x)g′(x)[g(x)]2=1x⋅x−ln⁡(x)⋅1x2.\frac{dy}{dx} = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2} = \frac{\frac{1}{x}\cdot x - \ln(x)\cdot 1}{x^2}.

Simplify the numerator: 1x⋅x=1\frac{1}{x}\cdot x = 1, so the numerator becomes 1−ln⁡(x)1 - \ln(x). Therefore dydx=1−ln⁡(x)x2.\frac{dy}{dx} = \frac{1 - \ln(x)}{x^2}.

Sign of the derivative's numerator term from the log rule

Hint
Step-by-step solution

The natural log term is subtracted. Recall the simplified result was one minus natural log of x, all over x squared. The quotient rule produced one times x minus natural log of x times one, and after simplifying, the log term keeps its negative sign in the numerator.

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