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

1 Answer
Jul 30, 2016

#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))#