Question

If f(x) = ln(x+e^-3x), then f '(0)=?

Answer 1

Use the chain rule to differentiate it, then plug in zero.

Answer 2

thanks im just not sure how to use the chain rule with this particular problem. i'm confused with the ln and e

Answer 3

Split them into seperate functions

outside function we will call s(x)
s(x) = ln(x) s'(x) = 1/x
inside function call g(x)
g(x) = x + e^(-3x) g'(x) = 1 + m'(x)
so we have (see below):
g'(x) 1 + -3e^(-3x)

so now

f'(x) = (1/(x + e^(-3x))) * (1 + -3e^(-3x))
f'(x) = (1 + -3e^(-3x))/(x + e^(-3x))

Solving m'(x)

We need to use chain rule on e^(-3x) to differentiate it so lets call it m(x)

lets call the outside function of o(x), m(x)

m(x) = e^(x) m'(x) = e^(x)
the exponent of e^(x) we will call l(x)
l(x) = -3x l'(x) = -3
so
m'(x) = -3e^(-3x)
so we have
g'(x) = 1 - 3e^(-3x)

Answer 4

This is the long way of solving chain rule.

conversely you can just stick to the rule

d/dx ln(f(x)) = f'(x)/f(x)

Answer 5

I recommend trying to understand my method if you do you will never have another problem differentiating again :)

Answer 6

Tell me if you need further help, for chain rule just remember


Where f(x) is your outside function and g(x) is your inside function

then you are just left with

f'(g(x)) * g'(x)