Rigoberto Mann

2023-03-07

What is the derivative of $y\uff1d\frac{\mathrm{ln}x}{x}$?

Nick Osborn

Beginner2023-03-08Added 3 answers

Use the quotient rule, which states that

$\frac{d}{dx}\left[\frac{f\left(x\right)}{g\left(x\right)}\right]=\frac{f\prime \left(x\right)g\left(x\right)-g\prime \left(x\right)f\left(x\right)}{{\left[g\left(x\right)\right]}^{2}}$

Applying this to $y=\frac{\mathrm{ln}x}{x}$, we see that

$y\prime =\frac{x\frac{d}{dx}\left[\mathrm{ln}x\right]-\mathrm{ln}x\frac{d}{dx}\left[x\right]}{{x}^{2}}$

Since $\frac{d}{dx}\left[\mathrm{ln}x\right]=\frac{1}{x}$ and $\frac{d}{dx}\left[x\right]=1$,

$y\prime =\frac{x\left(\frac{1}{x}\right)-\mathrm{ln}x}{{x}^{2}}$

$y\prime =\frac{1-\mathrm{ln}x}{{x}^{2}}$

$\frac{d}{dx}\left[\frac{f\left(x\right)}{g\left(x\right)}\right]=\frac{f\prime \left(x\right)g\left(x\right)-g\prime \left(x\right)f\left(x\right)}{{\left[g\left(x\right)\right]}^{2}}$

Applying this to $y=\frac{\mathrm{ln}x}{x}$, we see that

$y\prime =\frac{x\frac{d}{dx}\left[\mathrm{ln}x\right]-\mathrm{ln}x\frac{d}{dx}\left[x\right]}{{x}^{2}}$

Since $\frac{d}{dx}\left[\mathrm{ln}x\right]=\frac{1}{x}$ and $\frac{d}{dx}\left[x\right]=1$,

$y\prime =\frac{x\left(\frac{1}{x}\right)-\mathrm{ln}x}{{x}^{2}}$

$y\prime =\frac{1-\mathrm{ln}x}{{x}^{2}}$