Could someone check this simple differentiation?

2:50 Publicado por Mario Galarza

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Maths and statistics discussion, revision, exam and homework help.

1 Hour Ago: 17th October 2011 00:11

Could someone check this simple differentiation? Hi - using some online exam software - it's pretty rubbish and I'm trying to check my marks.

The question:

Differentiate

I get 4ln(5x) - 20-------------ln(5x)^2Thanks

Last edited by Geo877; 1 Hour Ago at 00:21.

1 Hour Ago: 17th October 2011 00:21

Re: Could someone check this simple differentiation? i do a level math (badly most of the time) but i dont think that is right at all. when i differentiate that, i end up with dy/dx = 2.458 (3d.p.)

when u differentiate u cancel all x's from what i can remember leavin u with dy/dx = 1/4(In5) (cant do fancy equation thing but its 4 over In5) then if u go on to solve for natural log of 5 nd divide 4 by ans

dont kno if this helps at all

WOOps forgot about the squaring dy/dx = 1.544 (3d.p.)

Last edited by ToH12; 1 Hour Ago at 00:24.

1 Hour Ago: 17th October 2011 00:46

Re: Could someone check this simple differentiation? Let me see.

I am just confused now. I always thought that:

$\dfrac{d}{dx}ln(f(x)) = \dfrac{f'(x)}{f(x)}$

But:

$\dfrac{d}{dx}ln(5x) = \dfrac{1}{x}$

I am just confused.

Last edited by Math-Illiterate; 1 Hour Ago at 00:54.

1 Hour Ago: 17th October 2011 00:55 Re: Could someone check this simple differentiation? i do a level math (badly most of the time) but i dont think that is right at all. when i differentiate that, i end up with dy/dx = 2.458 (3d.p.)

dont kno if this helps at all

WOOps forgot about the squaring dy/dx = 1.544 (3d.p.)

I, I, I... what? If you differentiate things, you should get a function and not a constant (the exception being if you differentiate something of the form cx, where c is a constant and x is your variable). That is not how you differentiate and I sincerely hope the OP ignores your post.

I don't do LaTeX (I need to learn!) but I get a fraction. The numerator:

4ln(5x) - (4/5)

The denominator:

(ln (5x) )^2

This isn't what you got. Pro tip: don't do questions like this online. Wolfram Integrator is trustworthy, but most other programs aren't, and it's best to learn to do it yourself anyway.

58 Minutes Ago: 17th October 2011 00:57 Re: Could someone check this simple differentiation? Let me see.

I am just confused now. I always thought that:

$\dfrac{d}{dx}ln(f(x)) = \dfrac{f'(x)}{f(x)}$

But:

$\dfrac{d}{dx}ln(5x) = \dfrac{1}{x}$

I am just confused.

Use the chain rule.

Let u = 5x, so du/dx = 5. Then d/du (ln u) = 1 / u = 1 / 5x. But multiplied by 5 this gives 5 / 5x, which simplifies to 1 / x.

... which means the differentiation I did quickly in that other post is wrong, because I didn't think about it. I give up; I'm going to bed. The point still stands, however, that the first post in this thread is pretty unhelpful!

54 Minutes Ago: 17th October 2011 01:01

Re: Could someone check this simple differentiation? Use the chain rule.

Let u = 5x, so du/dx = 5. Then d/du (ln u) = 1 / u = 1 / 5x. But multiplied by 5 this gives 5 / 5x, which simplifies to 1 / x.

I understand how you can differentiate to that. You can do it easier just by using the log rules:

It's super easy from there. It's just I thought that the rule I said earlier was how you did it. Ahh well.

52 Minutes Ago: 17th October 2011 01:03 Re: Could someone check this simple differentiation? I'm not sure what's going on but your answer is wrong. The correct answer is:

$\frac{4\ln(5x) - 4}{(\ln 5x)^2}$