Question

If f(x) = ln(1 − 5x), then find f '(x).

Answer 1

Can you please help? @123AB456C

Answer 2

Chain rule!

Answer 3

So the major rule of ln is
\[\ln(x)=\frac{ 1 }{ x }*(x)'\] remember that one?

Answer 4

here your
x=1-5x

Answer 5

Yay Baby's training in calc has paid off!

Answer 6

\[\ln(1-5x)=\frac{ 1 }{ 1-5x }*(1-5x)'\] right? :)

Answer 7

haha @Astrophysics I saw this question and I was like "YES. I UNDERSTAND IT" xD are you proud of me? ^v^ haha

Answer 8

Very xD

Answer 9

You taught me well :')

Oh @aryana_maria2323 It's important to note that that rule for ln is when you are doing it's derivative. not when you are simplifying it. That's a derivative rule :)

Answer 10

I tried using the chain rule but it didn't give me any of the answers.

Answer 11

Hm, try simplifying what I gave you. \[\ln(1-5x)=\frac{ 1 }{ 1-5x}*(1-5x)' = ?\]

Answer 12

\[f(g(x)) = f'(g(x))g'(x)\] chain rule

Answer 13

These are the answers

a. 5 over the quantity 5 times x minus 1
b. 1 over the quantity 1 minus 5x
c. 1 divided by x
d. 5 over the quantity 5 times x plus 1

Answer 14

I see the answer, hehe. When you simplify what i gave you, what do you get?

Answer 15

yes that is the chain rule I used. Okay let me simplify your problem @Babynini. One second.

Answer 16

\[\frac{ 1 }{ 1-5x }*(1-5x)'\]

You can ignore the first part (before the *) because that is already all derived. What does
(1-5x)' = ?

Answer 17

-5

Answer 18

Perfect, so now we have
\[\frac{ 1 }{ 1-5x }*-5\] so we multiply them and get \[\frac{ -5 }{ 1-5x }\]

Answer 19

so derive that?

Answer 20

Nope! That's already ln derived all the way.
\[\ln(1-5x)\]\[\frac{ d }{ dx }\ln(1-5x) = \frac{ 1 }{ 1-5x }*(1-5x)'\]\[\frac{ 1 }{ 1-5x }*-5\]\[\frac{ -5 }{ 1-5x }\] therefore\[\frac{ d }{ dx }\ln(1-5x)=\frac{ -5 }{1-5x }\]

Answer 21

Which simplifies to \[\frac{ 5 }{ 5x-1 }\]

Answer 22

By multiplying the top and bottom by -1 :) yaya good teamwork astro xD

Answer 23

\[\frac{ -5 }{ 1-5x }*\frac{ -1 }{ -1 }=\frac{ 5 }{ -1+5x }\]\[=\frac{ 5 }{ 5x-1 }\]

Answer 24

YAY!! Thank you guys so much. Can you help me with one more question?

Answer 25

For sure! put it on a new thread ^v^

Answer 26

Okay than you and I will tag you in it.