What is the derivative of #y=x^(2/x)#?
1 Answer
Jul 30, 2016
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
#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^'=(2x^(2/x)(1-ln(x)))/x^2#
Note that
#y^'=2x^(2/x-2)(1-ln(x))#