Question

Correct answer=brainliest

Answer 1

Explanation:

Let y = x^x^x

apply log on both sides

log y = x^x log x ---> equation 1

Now consider x^x

y = x^x

apply log on both sides

log y = x log x

1/y × dy/dx = 1 + xlogx

dy/dx = y(1+xlogx) -----> equation 2

diffentiate equation 1

1/y × dy/dx = d/dx(x^x) logx + 1/x (x^x)

so,

dy/dx = x^x^x( x^x(1+logx)logx + x^x-1)

Please check the answer once. Im not sure about this