Question

What is the derivative of `y=x^(2/x)`?

Answer 1

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

Explanation:

Use logarithmic differentiation. Start by taking the natural logarithm of both sides.

`ln(y)=ln(x^(2/x))`

The right hand side can be simplified using the logarithm rule `log(a^b)=blog(a)`.

`ln(y)=(2ln(x))/x`

Differentiate both sides. The chain rule will be necessary on the left-hand side, and the quotient rule on the right.

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

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

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

Now, solve for `y^'` by multiplying both sides by `y`. However, write `y` as `x^(2/x)` on the right.

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

Note that `x^(2/x)/x^2=x^(2/x-2)`. Thus:

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