Question

what is the derivative of x^(1/x)?? @MIT 18.02 Multiva…

Answer 1

Let y = x^(1/x). Then
ln y = 1/x . ln x
hence differentiating wrt x
1/y . y' = 1/x . 1/x + -1/x^2 . ln x

Hence

y' = 1/x^2 (1 - ln x) y
= 1/x^2 (1 - ln x) x^(1/x)

Answer 2

I.e.,
\[ \frac{d\ }{dx} x^{1/x} = \frac{1 - \ln x}{x^2} x^{1/x} \]

Answer 3

\[y=x^{\frac{1}{x}}\Longrightarrow\ln y = \frac{1}{x}\ln x\Longrightarrow \frac{1}{y}\cdot\frac{dy}{dx}=\left(-\frac{1}{x^2}\right)\ln x+\frac{1}{x}\cdot\frac{1}{x}\]\[\Longrightarrow\frac{dy}{dx}=y\left(\frac{1-\ln x}{x^2}\right)=x^{\frac{1}{x}}\left(\frac{1-\ln x}{x^2}\right)\]

Answer 4

ooo ok thx