This. . It's too difficult to write. . . Parentheses are also messy. Please copy them on the draft paper before you understand them. Next, I try to push it step by step.

Mainly use the formula (1+a) (1/a) = E A (the specific letter may be wrong, e is the natural logarithm, a→0).

[f(x+hx)/f(x)]^( 1/h)=e^( 1/x]

The left side of the equal sign is deformed {1+[f (x+hx)-f (x)]/f (x)} (1/h).

Then {1+[f (x+hx)-f (x)]/f (x)} {[f (x+hx)-f (x)]/[f (x+hx)-f (x)] * (65438+

Then, notice that the value f(x+hx)-f(x) in square brackets approaches 0 (because f(x) is derivable in (-∞,+∞)), f (x) >; 0), so use the formula (1+A) (1/A) = E A, and the left side of the equal sign is equal to

E [(f (x+hx)-f (x))/(h * f (x))] equals the equal sign, and the right side equals E (1/x).

Therefore, (f (x+hx)-f (x))/(h * f (x)) =1/x is deformed again.

(f(x+hx)-f(x))/h*x=f(x)/x^2

At this point, the left side of the equal sign is equal to f'(x), that is

f'(x)=f(x)/x^2

Add them together and you will know that f (x) = ce (- 1/x).