If f(x)=ln|(1-x)e^(-3x)|, please find f '(2)=?

We've got a few layers of chain rules and product rules. Let's try to break this down:

y = ln|(1 - x)e^(-3x)|

if we say:

u = |(1 - x)e^(-3x)|

then we have:

y = ln(u)

This derivative is simple:

dy/du = 1/u

Now we need to get du/dx, which is another chain rule:

u = |(1 - x)e^(-3x)|

if we say:

v = (1 - x)e^(-3x)

then:

u = |v|

du/dv = -1 if x > 1

du/dv = 1 if x < 1

Since we are looking for f'(2) ultimately, let's call this:

du/dv = -1

Now we need dv/dx.  Product rule:

v = (1 - x)e^(-3x)

dv/dx = (-1)e^(-3x) + (1 - x)(-3)e^(-3x)

dv/dx = -e^(-3x) - 3(1 - x)e^(-3x)

dv/dx = [-1 - 3(1 - x)]e^(-3x)

dv/dx = (-1 - 3 + 3x)e^(-3x)

dv/dx = (-4 + 3x)e^(-3x)

dy/dx = dy/du * du/dv * dv/dx

dy/dx = (1/u) * (-1) * (-4 + 3x)e^(-3x)

dy/dx = -(-4 + 3x)e^(-3x) / u

dy/dx = (4 - 3x)e^(-3x) / u

Put in u's expression back:

dy/dx = (4 - 3x)e^(-3x) / |(1 - x)e^(-3x)|

Now set x = 2 and solve for dy/dx:

dy/dx = (4 - 3 * 2)e^(-3 * 2) / |(1 - 2)e^(-3 * 2)|

dy/dx = (4 - 6)e^(-6) / |(-1)e^(-6)|

dy/dx = -2e^(-6) / e^(-6)

dy/dx = -2

So after all that:

f'(2) = -2

f(x)=ln|(1-x)e^(-3x)|. If 1-x<0 or x>1 then (1-x)e^(-3x)<0.

So write

f(x)=ln[-(1-x)e^(-3x)] (remove the | | sign first)=>

f '(x)=-[-3(1-x)e^(-3x)-e^(-3x)]/[-(1-x)e^(-3x)]=>

f '(x)=[e^(-3x)][3-3x+1]/[-(1-x)e^(-3x)]=>

f '(x)=[4-3x]/[-(1-x)]=>

f '(2)=[-2]/[1]=-2

Put f(x) = ln|g(x)| where g(x) = (1-x)e^(-3x).;

f'(x) = g'(x)/g(x)...(1).;

g'(x) = (-1)e^(-3x) + (-3)(1-x)e^(-3x) = (3x-4)e^(-3x).;

g(2) = (1-2)e^(-6) = -e^(-6).;

g'(2) = [3(2)-4]e^(-6) = 2e^(-6).;

(1)---> f'2) = g'(2)/g(2) = 2e^(-6)/-e^(-6) = -2.

f(x) = ln|(1-x)e^(-3x)|

since e^(-3x) ≥ 0 for all x

f(x) = ln(e^(-3x)|1-x|) = ln|1-x| + ln(e^(-3x)] 

 = ln|1-x| - 3x

Assuming real x:

f'(x) = 1/|1-x| - 3

f'(2) = 1/|1-2| -3 = 1 - 3 = -2

Ans: f'(2) = -2