Calc

[fossil hemlock]

Note that (ln(g))' = g'/g.

[drowsy locust]

is there a way to get g? is that what inverse derivatives are for. like could i use arctan(g)?

[fossil hemlock]

No need.
You know g(1) already, and you can calculate g'(1) by substituting x = 1 into g'(x). Then f'(1) = g'(1)/g(1).

[drowsy locust]

got it. still trying to understand it tho. the (ln(g)) =. is that a universal rule?

[fossil hemlock]

It's just an application of chain rule.
d(ln(g))/dx = (d(ln(g))/dg)(dg/dx) = (1/g)(g') = g'/g

[drowsy locust]

$d(ln(g))/dx = (d(ln(g))/dg)(dg/dx) = (1/g)(g') = g'/g$

[flat glade]

Flamester7 Tv

[drowsy locust]

oh ok ok

thank you