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Could someone check this simple differentiation?

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

1 Hour Ago: 17th October 2011 00:11 Geo877 0

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. Old

1 Hour Ago: 17th October 2011 00:21 ToH12 +1

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. Old

1 Hour Ago: 17th October 2011 00:46 Math-Illiterate +1

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. Old

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. :sigh: 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 Math-Illiterate +1

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:

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

I'll leave the working out to you - any specific questions, just ask.

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Could someone explain please?

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

33 Minutes Ago: 17th September 2011 21:00 zoxe +1

So i'm going through my C1 book and they say this:

x^2 = 9x
x^2 - 9x = 0
x(x-9) = 0

Then they say x = 0 or 9. I understand how x could be 9, and i understand why they factorised it, but i don't understand why x = 0, could someone explain?

Also, how would i go about doing this question?

3x^2 = 5?

Last edited by zoxe; 24 Minutes Ago at 21:08. Old

28 Minutes Ago: 17th September 2011 21:05 Re: Could someone explain please? So i'm going through my C1 book and they say this:

x^2 = 9x
x^2 - 9x = 0
x(x-9) = 0

Then they say x = 0 or 9. I understand how x could be 9, and i understand why they factorised it, but i don't understand why x = 0, could someone explain?

Also, how would i go about doing this question?

3x^2 = 5?

BASICALLY - for x(x-9) = 0. X can either be 9 which is obvious or 0 becuase 0 x 0 = 0 and 0 x -9 = 0 so x = 0. lol Last edited by kidoo; 21 Minutes Ago at 21:12. Old

24 Minutes Ago: 17th September 2011 21:08 Re: Could someone explain please? x(x-9) = 0
x=0/(x-9)
x=0

or x=9

3x^2=5
x^2=5/3
x=SQRT(5/3)

or x=-SQRT(5/3)

there you go armador, happy now?

Last edited by wannabeme; 12 Minutes Ago at 21:20. Old

20 Minutes Ago: 17th September 2011 21:12 Re: Could someone explain please? Its really simple. so the left hand side of the equation has to equal 0.

So seperate x(x-9)=0 to x=0 and x-9=0. Now rearrange to get x on its own...do this to both of them(x=0 is already done).

this will get you x=0 and x=9.

so either x=0 or x=9 to to make x(x-9)=0

16 Minutes Ago: 17th September 2011 21:17 Re: Could someone explain please? or x=9, which you conveniently glossed over when blindly dividing by something which could be null. OR x=-sqrt(5/3), but you also conveniently glossed over that by assuming that x^2 was an affine function, and therefore, only had one solution. Old

15 Minutes Ago: 17th September 2011 21:18 zoxe +1

Re: Could someone explain please? I'm still pretty unsure about this, like for this one i was solving:

x(6x+11) = 0
x = -11/6 <--- i could get this
x = 0 <--- still not sure why this is possible as in this equation:

2(x^2 - 4) = 0
x = 2
x = -2 and NOT 0

10 Minutes Ago: 17th September 2011 21:22 Re: Could someone explain please? or x=9, which you conveniently glossed over when blindly dividing by something which could be null.the question was how x(x-9) could equal 0, not what it equals, so I merely explained what they were asking for. OR x=-sqrt(5/3), but you also conveniently glossed over that by assuming that x^2 was an affine function, and therefore, only had one solution.Yeah this one is my fault, I'll go change it. Old

9 Minutes Ago: 17th September 2011 21:23 Re: Could someone explain please? I'm still pretty unsure about this, like for this one i was solving:

x(6x+11) = 0
x = -11/6 <--- i could get this
x = 0 <--- still not sure why this is possible as in this equation:

2(x^2 - 4) = 0
x = 2
x = -2 and NOT 0

The idea is this. Say you know that $a \times b = 0$ . Then either

, or both. (Because if neither of

and

were zero then clearly $a \times b$ couldn't be zero!)

So say instead of just and you have some expressions in . In your OP, you have and . Then using this reasoning we have that if then either or . But rearranging the equation just gives . Summarising this, we get that if then either or . We could say "or both" here, but is just a number so it can't be both 0 and 9!

I'll give you another slightly more complicated example. Say that . Then either or or . Rearranging gives that either or or $x=\frac{4}{3}$ . {Again, we could say "or any combination of the above", but can't take more than one value.}

In the equation , we can factorise to get . Thus either 2=0 (which can't happen, so we discard this) or (so ) or (so ), so we have that if then either or . There's no reason why the solution should appear here {and, indeed, it doesn't}.

We can't say with certainty which of the specific values takes, but you can say with certainty that takes one of those values (which is why we say "or").

Last edited by nuodai; 7 Minutes Ago at 21:26.

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