Question #81d94

1 Answer
Aug 27, 2016

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

Explanation:

First, use the rule that #(a^b)^c=a^((bc)# so #y=x^((x*x))=x^(x^2)#. We have:

#y=x^(x^2)#

Take the natural logarithm of both sides:

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

Simplifying this using #log(a^b)=blog(a)#:

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

Differentiate both sides. On the left side, the chain rule will kick into effect, since this is a function of #y# and not #x#. On the right hand side, we will use the product rule #(#if #y=fg#, then #y^'=f^'g+fg^')#.

We get:

#1/y*y^'=(x^2)^'ln(x)+x^2(ln(x))^'#

The derivative of #x^2# is #2x# and of #ln(x)# is #1/x#:

#1/y*y^'=2xln(x)+x^2(1/x)#

#1/y*y^'=x(2ln(x)+1)#

Now, solve for #y^'# by multiplying both sides of the equation by #y#.

#y^'=yx(2ln(x)+1)#

Since #y=x^(x^2)#:

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

Simplify #x^(x^2)x# using the rule #a^ba^c=a^(b+c)#:

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