How do you differentiate #y=x^lnx#?

1 Answer
Apr 18, 2016

Answer:

#y'=2x^(ln(x)-1)ln(x)#

Explanation:

Using logarithmic and implicit differentiation, take the natural logarithm of both sides.

#ln(y)=ln(x^ln(x))#

Simplify the right hand side using the rule: #ln(a^b)=b*ln(a)#

#ln(y)=ln(x)*ln(x)#

#ln(y)=(ln(x))^2#

Take the derivative of both sides. Recall that the chain rule is in effect on both sides.

#(y')/y=2ln(x)overbrace(d/dx(ln(x)))^(=1//x)#

#(y')/y=(2ln(x))/x#

Now, multiply both sides by #y=x^ln(x)# to solve for #y'#.

#y'=(2x^ln(x)ln(x))/x#

Note that we can simplify this be writing #x^ln(x)/x# as #x^ln(x)/x^1=x^(ln(x)-1)#. Thus,

#y'=2x^(ln(x)-1)ln(x)#