Question

Find the second derivative of y=x^x

Answer 1

Step-by-step explanation:

y=x^x=(e^1nx)^x=x .1n×e

y'all=e^x1n×x(x.1/x+(n×1)

y'all=x^x×(1+1nx)y"=x^x×(1/x)+(1+1nx)×x^

Answer 2

Answer:

y=X^x=(e^1nx)^x=x.1nxe

y'=e^x1nx × (X. 1/X+(nx1)

y'=X^x × (1+1nx)

y"=X^x × (1/X)+(1+1nx) ×X^x × (1+1nx)

y"=X^x [1/x+X(1+1nx)²/X]

y"=X^x × [1+X(1+1nx)²]/X] =0

=> y=X^x , x>o