Derivative of Log Function

**dark-ryder341** · Oct 31st 2009, 06:57 PM

Hey guys,

I'm really having some difficulty understanding how to find the derivative of log functions - I keep trying these homework questions and getting most of them wrong and I'm really starting to get annoyed.

Here's a question, for example:

y=ln(4x)/x^4
y'=?

I did attempt it several times, different ways but each time it was wrong...

Any hints on how to go about these sorts of questions? I know that you're supposed to use the chain rule but it's just not working for me for some reason.

Any help's appreciated!

**Mush** · Oct 31st 2009, 07:08 PM

Originally Posted by dark-ryder341

Hey guys,

I'm really having some difficulty understanding how to find the derivative of log functions - I keep trying these homework questions and getting most of them wrong and I'm really starting to get annoyed.

Here's a question, for example:

y=ln(4x)/x^4
y'=?

I did attempt it several times, different ways but each time it was wrong...

Any hints on how to go about these sorts of questions? I know that you're supposed to use the chain rule but it's just not working for me for some reason.

Any help's appreciated!

Well you have to use the quotient rule on this one first of all. Followed by the chain rule.

**dark-ryder341** · Oct 31st 2009, 07:21 PM

Alright...so, I'd go:

ln(4x)' * x^4 - ln(4x) * x^4' / (x^4)^2

1/4x * x^4 - ln(4x) * 4x^3 / (x^4)^2

x^4/4x - ln(4x) * 4x^3 / (x^8) [I think you would change it to x^8?]

I get stuck at this point though...would I just leave it like that and then apply the chain rule?

**Mush** · Oct 31st 2009, 07:29 PM

Originally Posted by dark-ryder341

Alright...so, I'd go:

ln(4x)' * x^4 - ln(4x) * x^4' / (x^4)^2

1/4x * x^4 - ln(4x) * 4x^3 / (x^4)^2

x^4/4x - ln(4x) * 4x^3 / (x^8) [I think you would change it to x^8?]

I get stuck at this point though...would I just leave it like that and then apply the chain rule?

Very close, but remember the chain rule.

$\frac{d}{dx} \ln(4x) = \frac{1}{4x} \times \frac{d}{dx} (4x) = \frac{1}{4x} \times 4 = \frac{1}{x}$ , so your result should be:

$\frac{\frac{x^4}{x} - \ln(4x) \times 4x^3}{x^8}$

Which simplifies to $\frac{x^3 - \ln(4x) \times 4x^3}{x^8}$

Which simplifies even further to $\frac{1 - 4 \ln(4x)}{x^5}$ by taking out a factor of $x^3$ on the top and bottom.

**dark-ryder341** · Oct 31st 2009, 07:34 PM

Alright, so I'd get:

x^3/4 - ln(4x) * 4 / x^5, then? Would I be able to take the x^3 out of that fraction as well, though? Thanks for helping me by the way.

**Mush** · Oct 31st 2009, 07:37 PM

Originally Posted by dark-ryder341

Alright, so I'd get:

x^3/4 - ln(4x) * 4 / x^5, then? Would I be able to take the x^3 out of that fraction as well, though? Thanks for helping me by the way.

I noticed a mistake, and I edit my above post to reflect this. Have a look.

**dark-ryder341** · Oct 31st 2009, 08:03 PM

Ah, I see! Thanks. So I wouldn't have to reduce that any more, right? Since the chain rule was already applied to ln(4x)...? Would it be the final answer?

**Mush** · Oct 31st 2009, 08:32 PM

Originally Posted by dark-ryder341

Ah, I see! Thanks. So I wouldn't have to reduce that any more, right? Since the chain rule was already applied to ln(4x)...? Would it be the final answer?

Yep, that would be the final simplified answer.

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