The chain rule and ln...

Hey Guys, another question...

if you had to differentiate ln3x you would assume you could do it like this:

ln3x = ln3 + lnx

so you will just get 1/x as ln3 is a constant

Howevere... If you did the chain rule and made u = 3x you would get 3/x as your answer, but this is wrong.

However some questions like - differentiate ln (3x + 1) you have to use the chain rule, I don't understand how you can use the chain rule here and get a correct answer, but if you used in the first instance you get the wrong answer.

Thanks :smile:

Reply 1

Potassium^2

y=ln(u) where u=3x

\frac{dy}{dx}=\frac{dy}{du} \cdot \frac{du}{dx}

=\frac{1}{u} \cdot 3

=3/3x , 3's cancel
=1/x

(edited 13 years ago)

Reply 2

jaheen22

12

You can differentiate ln(3x) different ways. Each way will produce 1/x and some constant. The constants are all different however so you actually haven't done it wrong.

Reply 3

harr

15

Original post by eddhann

However some questions like - differentiate ln (3x + 1) you have to use the chain rule

ln(3x+1)=ln(3(x+1/3)) is just a translation of ln(3x), so you can work out the derivative of ln(3x) and translate it rather than using the chain rule. It's probably easier to use the chain rule, but you don't have to.

Original post by jaheen22

You can differentiate ln(3x) different ways. Each way will produce 1/x and some constant. The constants are all different however so you actually haven't done it wrong.

?

(edited 13 years ago)

Reply 4

ViralRiver

16

Original post by eddhann

Hey Guys, another question...

if you had to differentiate ln3x you would assume you could do it like this:

ln3x = ln3 + lnx

so you will just get 1/x as ln3 is a constant

Howevere... If you did the chain rule and made u = 3x you would get 3/x as your answer, but this is wrong.

However some questions like - differentiate ln (3x + 1) you have to use the chain rule, I don't understand how you can use the chain rule here and get a correct answer, but if you used in the first instance you get the wrong answer.

Thanks :smile:

It seems you're not understanding the chain rule. You're correct in saying that

ln3x = ln3 + lnx

and hence

\frac{d}{dx}ln3x=\frac{1}{x}

.

In order to do the chain rule, do the following:

\text{let } y = ln3x, u = 3x\Rightarrow y = lnu

Then differentiate according to the chain rule:

\frac{du}{dx}=3, \frac{dy}{du}=\frac{1}{u}

\frac{dy}{dx}=\frac{dy}{du}\cdot \frac{du}{dx}=3\cdot \frac{1}{u}=\frac{3}{u}=\frac{3}{3x}=\frac{1}{x}

Reply 5

limetang

19

For this we can say 3x=f(x). Now for ln (x) you can basically say dy/dx = f`(x)/f(x). So dy/dx= 1/x.

Or for using the chain rule we can say this let u=3x du/dx=3 and then say y=ln(u) so dy/du = 1/u. Now since dy/dx = du/dx * dy/du we can say dy/dx= 3/u =3/3x =1/x.

Reply 6

jaheen22

12

Original post by harr

?

I'm on drugs :cool:

Reply 7

eddhann

OP

10

thanks everyone :biggrin:

The chain rule and ln...

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