What is the dervative of x pwr x pwr x?

Answer 1

For the function:

y = x^x^x (the superscript notation on this text editor does not work with double superscripts)

To solve for the derivative y', implicit differentiation is needed. First, the equation must be manipulated so there are no x's raised to x's on the right side of the equation. So, both sides of the equation must be input into a natural logarithm, wherein we can use the properties of logarithms to remove the superscripted powers of the right side:

ln(y) = ln(x^x^x)

ln(y) = xxln(x)

ln(y)/ln(x) = xx

ln(ln(y)/ln(x)) = xln(x)

eln(ln(y)/ln(x)) = exln(x)

ln(y)/ln(x) = exln(x)

ln(y) = ln(x)exln(x)

Now there are no functions raised to functions (x's raised to x's). Deriving this equation yields:

(1/y)(y') = ln(x)exln(x)(x(1/x) + ln(x)) + exln(x)(1/x) = ln(x)exln(x)(1 + ln(x)) + exln(x)(1/x) = exln(x)(ln(x)(1+ln(x)) + (1/x))

Solving for y' yields:

y' = y[exln(x)(ln2(x) + ln(x) + (1/x))]

or

y = xx^x

ln(y) = ln(x)x^x

ln(y) = xxln(x)

ln(y) = exlnxln(x)

y'/y = exlnx[ln(x) + 1)ln(x) + exlnx(1/x)

y' = y[exlnx(ln2(x) + ln(x) + 1/x)]

y' = xx^x[exlnx(ln2(x) + ln(x) + 1/x)]