Question #81d94
1 Answer
Explanation:
First, use the rule that
#y=x^(x^2)#
Take the natural logarithm of both sides:
#ln(y)=ln(x^(x^2))#
Simplifying this using
#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
We get:
#1/y*y^'=(x^2)^'ln(x)+x^2(ln(x))^'#
The derivative of
#1/y*y^'=2xln(x)+x^2(1/x)#
#1/y*y^'=x(2ln(x)+1)#
Now, solve for
#y^'=yx(2ln(x)+1)#
Since
#y^'=x^(x^2)x(2ln(x)+1)#
Simplify
#y^'=x^(x^2+1)(2ln(x)+1)#