Derivative Of Ln 3x 1

Because Maths sucks.

EDIT: Holyy craaaaaaaaaaaaapppppppp!!!!!!!!!!!!!!!!!!!!!!

NEGGED FOR Maxim MATHS SUCKS.

WOW.

d/dx ln ( f(x) ) = f'(x)/f(10)

So...

d/dx ln 3x = three/3x = 1/x

(Original post by Kasc)
d/dx ln ( f(x) ) = f'(x)/f(x)

So...

d/dx ln 3x = iii/3x = ane/10

where did you learn this?
???

During my A Levels?

what i mean was which chapter of the core books.

Using the chain dominion, dy/dx = 1/3x x 3. You know that ln x is differentiated to 1/x so ln 3x would be differentiated to 1/3x multiplied by the differential of what's in the bracket.

So one/3x 10 3 = three/3x. The 3s cancel out leaving you with i/10.

I did my A Level in maths 2 years agone, likewise we might non even have used/using the same text books!

this

and the detivative of this is only 1/x

The best way to differentiate natural logarithms is by knowing this pattern:

y = ln f(ten)

dy/dx = f'(x) / f(x)

If yous're using the C3 volume by Keith Pledger, it should be around the latter part of last the chapter.

Dorsum to question.

y = ln f(x)
y = ln (3x)

So f(x) = 3x and f'(ten) = 3

dy/dx = f'(x) / f(x)
dy/dx = 3 / 3x = 1/x (as y'all can see, the 3/iii will cancel out)

This would likewise assist:

The differential of ln iii is aught and the differential of ln(10) is ane/10.

Goodluck

(Original postal service past Blastoff-Omega)
The all-time way to differentiate natural logarithms is by knowing this pattern:

y = ln f(x)

dy/dx = f'(x) / f(10)

If you're using the C3 book by Keith Pledger, information technology should be effectually the latter office of concluding the chapter.

Back to question.

y = ln f(ten)
y = ln (3x)

So f(ten) = 3x and f'(x) = 3

dy/dx = f'(x) / f(10)
dy/dx = three / 3x = 1/x (as you lot can come across, the iii/3 will cancel out)

This would too help:

The differential of ln three is zero and the differential of ln(x) is 1/x.

Goodluck

Yep it is, why on earth was this a chapter 4 question, when they solved it using a chapter eight method.

To be honest guys, have a look at the OP; throwing all these methods at them is just going to confuse.

differentiate ln3x with regards to 3x and multiply by differential of 3x with regards to x
d(ln3x)/d(3x) times d(3x)/dx
giving you i/3x times 3= 3/3x=1/x
chain rule bro

Im not certain how to differentiate natural logs atm, which is why i had trouble.

Derivative Of Ln 3x 1,

Source: https://www.thestudentroom.co.uk/showthread.php?t=1671654

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