Urgent! Let f (b) = B LNA - a LNB (b > a > e), then what is f '(b) equal to? Given that a and B are real numbers, and a > b > e, where e is the base of natural logarithm, we prove that a ^ b > b ^ a

Urgent! Let f (b) = B LNA - a LNB (b > a > e), then what is f '(b) equal to? Given that a and B are real numbers, and a > b > e, where e is the base of natural logarithm, we prove that a ^ b > b ^ a

If f (b) = blna - AlNb, then B is an independent variable and a is a constant
f'(b)=lna -a/b
Note: if we want to prove blna > AlNb, (b > a > e), it can be changed into (LNA) / a > ln (b) / b
Let g (x) = (LNX) / x, x > e, then G '(x) = (1-lnx) / X & # 178; G (b)

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