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

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Answer 1 · 2016-07-30T17:02:16

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

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:

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

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