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# Maths Help Needed! :( watch

1. Hey everyone!

I'm stuck on a few questions we did in school. Could someone please help me on the following:

[list=1][*]Differentiate the following with respect to x: y= e^x ln x[*]Differentiate the following with respect to x: y= ln x/e^x[*]Could someone please tell me how you get from:

16(16-2x)-(16-2x)2x-16=0

to

(16-2x)(16-2x)=16

I would be really grateful if you could include the steps!

As you can see I'm not great at maths.

2. (Original post by Bilal786)

Differentiate the following with respect to x: y= e^x ln x
Use the chain rule. This entails seeing that the function is the product of two other functions, i.e. y = (e^x)(lnx).

In words: You write the first function and multiply it by the derivative of the second, then add to this the second function multiplied by the derivative of the first.

Thus, dy/dx = (e^x)(1/x) + (lnx)(e^x)

(Original post by Bilal786)
Differentiate the following with respect to x: y= ln x/e^x
This uses the quotient rule, which is essentially the chain rule, but with the second function raised to -1 (you can prove the standard formula like this).

In words:
- write the function on the bottom
- multiply this by the derivative of the function on the top
- subtract from this the function on the top multiplied by the derivative of the function on the bottom
- divide the whole thing by the function on the bottom squared

Thus, dy/dx = [(e^x)(1/x) - (lnx)(e^x)]/(e^2x)

No.3 can probably be done by factorising, and no.4 uses the quotient rule again.

These are both practical ways of applying the rules - if you're not sure how to differentiate such functions in the first place, you might not have covered them yet. I'd recommend looking up the rules to see where they come from, rather than just applying a 'work through' method, if you're not really sure what's going on.
3. (Original post by Bilal786)
Hey everyone!

I'm stuck on a few questions we did in school. Could someone please help me on the following:

1. Differentiate the following with respect to x: y= e^x ln x
2. Differentiate the following with respect to x: y= ln x/e^x
3. Could someone please tell me how you get from:

16(16-2x)-(16-2x)2x-16=0

to

(16-2x)(16-2x)=16

I would be really grateful if you could include the steps!
4. Differentiate the following with respect to x and find the value of x that gives a maximum: S= 8x^2 ln(1/2x)

As you can see I'm not great at maths.

what type of questions are these,p2,p1,p3??!!!!!!
4. Hi,

habosh these questions are from the Advanced Higher sylabbus which is the scottish equivalent to A-Levels.

These are the asnwers to the questions I posted:
1. dy/dx= e^x/x(x lnx + 1)
2. dy/dx= (1-x lnx)/(x e^x)
3. 0.303 (Question 4)

It's just getting to these answers which I'm stuck on.
5. Anyone?
6. (Original post by Bilal786)
Anyone?
Those answers are wrong. There must be a mistake in the book's printing (or I've forgotten what calculus I knew in the past two months).

See if the answers to the next set of questions or excercise actually match, as the two could be muddled.

Or, have I read the functions incorrectly? e.g. no.1 could be y = e^(xlnx). But that still doesn't differentiate to give your answer...
7. (Original post by Bilal786)
Anyone?
(1) We can write Werther's answer (e^x)(1/x) + (ln(x))(e^x) as

(1/x) * e^x * (x ln(x) + 1).

(2) We can simplify Werther's answer [(e^x)(1/x) - (ln(x))(e^x)]/(e^(2x)) by replacing e^(2x) by (e^x)(e^x) and then cancelling e^x:

[(1/x) - ln(x)]/e^x
= [1 - x ln(x)]/(x e^x).

(4) If S = 8x^2 ln(1/(2x)) then, using the rules for logs, S = -8x^2 ln(2x). The domain of S is x > 0. By the product rule,

dS/dx
= -8x^2 (1/x) - 16x ln(2x)
= -8x - 16x ln(2x)
= -8x (1 + 2 ln(2x)).

At a stationary point of S we have -8x (1 + 2 ln(2x)) = 0. Since x > 0,

1 + 2 ln(2x) = 0
ie, ln(2x) = -1/2
ie, 2x = e^(-1/2)
ie, x = (1/2)e^(-1/2)
ie, x = 0.303 (3dp).
Attached Images

8. Hey thanks for the help every1!!!!!!

I'm still stuck on Question 3 though.

Anyway, thanks for the help.

Much appreciated.
9. 16(16 - 2x) - (16 - 2x)2x - 16 = 0
16(16 - 2x) - (16 - 2x)2x = 16 . . . by adding 16 to both sides
(16 - 2x)16 - (16 - 2x)2x = 16 . . . because A*B = B*A
(16 - 2x)(16 - 2x) = 16 . . . factorising.
10. (Original post by Jonny W)
16(16 - 2x) - (16 - 2x)2x - 16 = 0
16(16 - 2x) - (16 - 2x)2x = 16 . . . by adding 16 to both sides
(16 - 2x)16 - (16 - 2x)2x = 16 . . . because A*B = B*A
(16 - 2x)(16 - 2x) = 16 . . . factorising.
I solve it the same,but got confused on
. (16 - 2x)16 - (16 - 2x)2x = 16
I did multiply the no ie 2x with it's bracket so it didn't work for me,would you please be so kind and explain to me how did you solve that bit,because 16 and 2x and very different numbers,so you cant just cancel them
11. You have:

(16 - 2x)16 - (16 - 2x)2x = 16

so what I'd do is say that (on the left hand side), you have a common factor of (16-2x) is each term, so:

(16-2x)16 - (16-2x)2x = (16-2x)(16-2x) = 16

(3x)8 - (3x)2, you could take out a factor of 3x to give 3x(8-2), which is 18x (ie 24x - 6x).
12. (Original post by 85ah11)
You have:

(16 - 2x)16 - (16 - 2x)2x = 16

so what I'd do is say that (on the left hand side), you have a common factor of (16-2x) is each term, so:

(16-2x)16 - (16-2x)2x = (16-2x)(16-2x) = 16

(3x)8 - (3x)2, you could take out a factor of 3x to give 3x(8-2), which is 18x (ie 24x - 6x).
Thanks alot,understood it,finally wooooooooooooohooooooo,thanx again

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