Question

Given the function \[x_1^{x_2^{x_3}}\]. Take the derivative of \[x_1, x_2, x_3\]

\[f \prime_{x_1}x = x^{x_2x_3}
\rightarrow x_2x_3*x_1^{(x_2x_3)-1}\]

I don't understand how to take the derivative of \[x_2\] and \[x_3\].. Some guidelines would be appreciated. :D

Answer 1

do u know chain rule ?

Answer 2

\[f(g(x)) = f\prime(g(x))*g\prime x\]

Answer 3

yes, [for diff. w.r.t x2]
so, x1 and x3 are constants and x2 is variable.
for simplicity, i'll take the function as \(a^{x^b}\)
using chain rule,
it'll be
\((a^{x^b})' = (a^{x^b})\ln x^b(x^b)'\)
got this ?

Answer 4

a =x1, x=x2, b=x3

Answer 5

doubts ?

Answer 6

its easier for diff. w.r.t x3
x1 and x2 are constants and x3 is variable.
for simplicity, i'll take the function as \(a^{b^x} \)
let a^b = c a new constant.
then you'll just need to differentiate c^x
substitute back in c = a^b
a=x1,b=x2,c=x3

Answer 7

*x=x3

Answer 8

Been scribbling on my own here and I guess I got it.

Answer 9

GOOD :)