The instantaneous rate of change of a function f(x) at a point where x = a is equal to f'(a), where f'(x) is the first derivative of f(x) with respect to x.

For the function `f(x) = (ln(x))/x^5` , determine f'(x) using the quotient rule.

`f'(x) = ((ln(x))'*x^5 - ln(x)*(x^5)')/(x^5)^2`

= `((1/x)*x^5 - ln(x)*5*x^4)/x^10`

= `(x^4 - ln(x)*5*x^4)/x^10`

= `(1 - ln(x)*5)/x^6`

At the point (1, 0), x = 1 and `f'(1) = (1 - ln(1)*5)/1^6 = (1 - 0)/1 = 1`

**The instantaneous rate of change of `f(x) = (ln(x))/x^5` at the point (1,0) is 1.**

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