# How do you differentiate y=x^(2/x)?

Nov 17, 2016

## $- \frac{4}{x} ^ 3 {x}^{\frac{2 - x}{x}}$

#### Explanation:

Apply Chain rule of differentiation, with $u = \frac{2}{x}$:
$y ' = \left({x}^{u}\right) ' \setminus \times \left(\frac{2}{x}\right) '$
set differentiation $\setminus \Rightarrow \left(u {x}^{u - 1}\right) \setminus \times - 2 {\left(x\right)}^{- 2}$
apply u=(2/x): $\setminus \Rightarrow \left(\frac{2}{x}\right) {x}^{\frac{2}{x} - 1} \setminus \times - 2 {x}^{- 2}$
fraction form of $- 2 {x}^{-} 2$: $\setminus \Rightarrow - \left(\frac{2}{x}\right) {x}^{\frac{2 - x}{x}} \frac{2}{x} ^ 2$

• the two fractions can multiply by each other.

## $- \frac{4}{x} ^ 3 {x}^{\frac{2 - x}{x}}$

Nov 17, 2016

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

#### Explanation:

As $y = {x}^{\frac{2}{x}}$

we have $\ln y = \frac{2}{x} \ln x$

or $2 \ln x = x \ln y$

Now deriving both sides w.r.t. $x$

$\frac{2}{x} = \ln y + \frac{x}{y} \cdot \frac{\mathrm{dy}}{\mathrm{dx}}$

or $\frac{\mathrm{dy}}{\mathrm{dx}} = \left(\frac{2}{x} - \ln y\right) \cdot \frac{y}{x}$

= $\left(\frac{2}{x} - \frac{2}{x} \ln x\right) \cdot \frac{y}{x}$

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

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