Home > Calculus > Differentiating Exponential Functions with Other Bases

What is the derivative of #x^x^x#?

Answer 1

Paisley Johnson

Since the rule:

#[f(x)]^g(x)=e^log([f(x)]^g(x))=e^(g(x)logf(x))#,

then:

#x^(x^x)=e^log(x^(x^x))=e^(x^xlogx)=e^(e^log(x^x)logx)=#

#=e^((e^(xlogx))logx)#.

So the function to derive is:

#y=e^((e^(xlogx))logx)#.

#y'=e^((e^(xlogx))logx)*[e^(xlogx)(1*logx+x*1/x)*logx+e^(xlogx)*1/x]=#

#=e^((e^(xlogx))logx)*e^(xlogx)(logx+1+1/x)=#

#=e^((e^(xlogx))logx)*e^(xlogx)(xlogx+x+1)/x#

And, if you want:

#y'=x^(x^x)*x^x(xlogx+x+1)/x#.

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Emma Aldridge

The derivative of (x^{x^x}) with respect to (x) is (x^{x^x} \cdot (1 + \ln(x) + \ln(x) \cdot x^x)).

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.