What is the derivative of $x^{\frac{1}{x}}$ $x^(1/x)$ ?

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What is the derivative of $x^{\frac{1}{x}}$ $x^(1/x)$ ?

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Answer 1 · 2018-04-02T18:37:08

$\frac{d y}{d x} = x^{\frac{1}{x}} (\frac{1 - ln x}{x^{2}})$ $dy/dx=x^(1/x)((1-lnx)/x^2)$

Explanation:

In these situations where a function is raised to the power of a function, we'll use logarithmic differentiation and implicit differentiation as follows:

$y = x^{\frac{1}{x}}$ $y=x^(1/x)$

$ln y = ln (x^{\frac{1}{x}})$ $lny=ln(x^(1/x))$

From the fact that $ln (a^{b}) = b ln a$ $ln(a^b)=blna$ :

$ln y = \frac{ln x}{x}$ $lny=lnx/x$

Differentiate (the left side will be differentiated implicitly):

$\frac{1}{y} \cdot \frac{d y}{d x} = \frac{1 - ln x}{x^{2}}$ $1/y*dy/dx=(1-lnx)/x^2$

Solve for $\frac{d y}{d x}$ $dy/dx$ :

$\frac{d y}{d x} = y (\frac{1 - ln x}{x^{2}})$ $dy/dx=y((1-lnx)/x^2)$

Recalling that $y = x^{\frac{1}{x}}$ $y=x^(1/x)$ :

$\frac{d y}{d x} = x^{\frac{1}{x}} (\frac{1 - ln x}{x^{2}})$ $dy/dx=x^(1/x)((1-lnx)/x^2)$

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