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.
See eNotes Ad-Free
Start your 48-hour free trial to get access to more than 30,000 additional guides and more than 350,000 Homework Help questions answered by our experts.
Already a member? Log in here.