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Brilliant. Thank you for showing me this JeffM. I followed this perfectly. Just need to practice more. Many thanks to all.

Ok. I'm not getting this. Using a*(b+c) how does this relate to the textbook solution? dy/dx = x2[6(2x + 1)2] + (2x + 1)3(2x) Common factor = 2x(2x + 1)2 dy/dx = 2x(2x + 1)2 [3x + (2x + 1)] dy/dx = 2x(2x + 1)2 (5x + 1)] Is it possible someone can show my step by step?

I remember, so to revise: 1st term to (x)6x[(2x + 1)2]. Then 6x[(2x + 1)2] divided by 2x(2x + 1)2 goes in three times, just leaving 2x(2x + 1)2 + 3x So 2x(2x + 1)2 + 3x + (2x + 1). Thus 2x(2x + 1)2 + 5(x + 1). Correct process?

Ok. I follow that Denis. But how to get the textbook answer (so I can understand his process)?

dy/dx = x2[6(2x + 1)2] + (2x + 1)3(2x) becomes dy/dx = 6x2[(2x + 1)2] + (2x)(2x + 1)3 Ok, I see 2x(2x + 1)2 goes into the 2nd term, leaving (2x + 1) So, now 2x(2x + 1)2 into the 1st term of 6x2[(2x + 1)2] . I'm not following here....I know this is going to be wrong, but just so you can see my...

Yes....it's the 3x and 2x outside the brackets that is most confusing.

Hi all, I'm learning the rules of differentiation, specifically the product rule. I understand the steps to obtain an answer, but am stuck when it comes to simplifying and taking out a common factor. For example, the answer to one of the problems is: dy/dx = x2[6(2x + 1)2] + (2x + 1)3(2x) I...

Ah yes. I understand I should have done that. Thanks for point this out.

Thanks mmm4444bot. So much easier with x(x+1). Don't know why I didn't see that. 5/x(x+1) - 2(x+1)/x(x+1) + 3x/x(x+1) 5 - 2x - 2 + 3x = x+3 Thus, x+3/x(x+1) Many thanks.

Question: . . . . .\mbox{(e) }\, \dfrac{5}{x\, (x\, +\, 1)}\, -\, \dfrac{2}{x}\, +\, \dfrac{3}{x\, +\, 1} I'm getting myself confused with this one. 1) I multiplied all the denominators, so they all became x(x+1)x(x+1). 2) Then 5x(x+1) - 2x(x+1)(x+1) + 3x(x+1)x 3) Becomes, 5x2 + 5x - 2x3 -...

Thanks Dennis. I hate when I thought I had tried every conceivable way, and yet that's one more I didn't try! Got it now. Thxs.

Q. During the next 3 years, a business decides to invest £10,000, at the beginning of each year. The revenue at the end of each year is: Year 1 = £5,000 Year 2 = £20,000 Year 3 = £50,000 Calculate the NPV using a discount rate of 4%, compounded annually. A. My answer was £39,997.85...

Thank you tkhunny. I calculated the correct answer adding up the individual payments.

Hi all, Q. A regular savings of $500 is made into a sinking fund at the start of each year for 10 years. Determine the value of the fund at the end of the tenth year on the assumption that the rate of interest is 10% compounded continuously. This is part b) of the question. I previously...

Ah right. Because when I expand it out: 101 x (1.01n-1) / 0.01 Then rearrange to: 101 / 0.01 x (1.01n-1) Becomes: 10100 x (1.01n-1) i.e. 10100 (1.01n-1) Great - thank you Dr.Peterson.

Morning/afternoon/evening all, I'm aware of the formula: a(rn-1/r-1). However, for the following: 100(1.01)(1.01n-1/1.01-1) Apparently (Mathematics for Economics and Business - Ian Jacques), this is the same as: 10,100(1.01n-1) Is that correct? I can't find out how to get to this...

I see where I went wrong now. When I multiplied both sides by 0.85, I didn't multiply the whole function, i.e. -5Q + 80 x 0.85, whereas it should have been 0.85(-5Q + 80). Thank you JeffM. Much appreciated.

Hi all, Trying to solve the following problem from Mathematics for Economics and Business by Ian Jacques - Supply and Demand Analysis. The demand and supply functions of a good are given by P = -5Q + 80 P = 2Q + 10 For this first part I've already worked out the equilibrium price and quantity...

Thank you, but to understand the formula? S30 = 30/2 * [a1 + an] S30 = 15 * [116] S30 = 1740 Great - thanks, that tallies up with the answer key, but how do you get from: Sn= n/2 * [2(a)+(n-1)d] to Sn= n/2 * [a1 + an] ?

Is the question wrong? Attached..