# Math Help - Derivative of Log Function

1. ## Derivative of Log Function

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!

2. 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.

3. 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?

4. 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.

5. 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.

6. 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.

7. 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?

8. 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.