Find y'(2)

[Lorena_Santoro]

https://youtu.be/1fdk5sWG9gI

[Guest]

I hate having to go to another site to read the problem. Especially when it is so simple to write.

We are given that y= x^x and want to find y'.

Take the logarithm of both sides: log(y)= log(x^x)= x log(x).
Now differentiate. (log(y))'= (1/y)y'= log(x)+ x/x= log(x)+ 1.
y'= y(log(x)+ 1)= x^x(log(x)+ 1).

y'(2)= 2^2(log(2)+ 1)= 4(log(2)+ 1).