Chain Rule Question

[Pudge714]

Could someone please help me derive the follwing two functions.

f(x)= xln(ln(x))

and

x^(e^x)

All help would be greatly appreciated.

[cts]

I suppose you mean differentiate?

f(x) = xlog(log(x))
f'(x) = x'log(log(x)) + x(log(log(x))'
= log(log(x)) + x(1/log(x))*1/x
= log(log(x)) + 1/log(x)

g(x) = x^(e^x)
g'(x) = x^(e^x)(log(x)e^x + e^x/x)

[Lestat]

[ QUOTE ]
I suppose you mean differentiate?

f(x) = xlog(log(x))
f'(x) = x'log(log(x)) + x(log(log(x))'
= log(log(x)) + x(1/log(x))*1/x
= log(log(x)) + 1/log(x)

g(x) = x^(e^x)
g'(x) = x^(e^x)(log(x)e^x + e^x/x)

[/ QUOTE ]

It's almost scary to me that there are people who can look at this and actually make sense of it.

[PokerPadawan]

Standard.

[tomdemaine]

[ QUOTE ]
[ QUOTE ]
I suppose you mean differentiate?

f(x) = xlog(log(x))
f'(x) = x'log(log(x)) + x(log(log(x))'
= log(log(x)) + x(1/log(x))*1/x
= log(log(x)) + 1/log(x)

g(x) = x^(e^x)
g'(x) = x^(e^x)(log(x)e^x + e^x/x)

[/ QUOTE ]

It's almost scary to me that there are people who can look at this and actually make sense of it.

[/ QUOTE ]

I think this is the formula for the new pattern mapper tm 2.0

[Meromorphic]

[ QUOTE ]
[ QUOTE ]
I suppose you mean differentiate?

f(x) = xlog(log(x))
f'(x) = x'log(log(x)) + x(log(log(x))'
= log(log(x)) + x(1/log(x))*1/x
= log(log(x)) + 1/log(x)

g(x) = x^(e^x)
g'(x) = x^(e^x)(log(x)e^x + e^x/x)

[/ QUOTE ]

It's almost scary to me that there are people who can look at this and actually make sense of it.

[/ QUOTE ]

I'm sure most of the people in freshman calculus could make similar comments about cryptic poker comments. Just a matter of learning what everything means.