[Closed] Anyone do maths? (again..)

How many thousandths in an inch ?
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I dont know, there must be millions of them!

or as some tool i went to school with once said while watching superbowl:

"how many quarters are there?"

6.

No clever people on today then.. 🙄

Once painfully witnessed two girl shop assistants trying to work out 15% of £1. Took them about 5 minutes, and they still had their doubts.

Does is not go like this

y = uv
dy/dx - u.dv/dx + v.du/dx

Now y = (x^2).(e^3x)

So, dy/dx = (x^2).(3.e^3x) + (2x).(e^3x) { I think the dy/dx of y = e^x = e^x and if y = e^3x then dy/dx = 3.e^3x}

= (e^3x).(3.x^2 + 2x)
= x.e^3x.(3x+2)
Substituting x = 1 gives
dy/dx(1) = 1.e^3.(5x1 + 2)
= 5.e^3

Then the y = mx + c bit gives the same answer as you. So I agree you are correct.

Gary PhD, MSc, BSc :roll:. I am probably completely wrong its close on 25 years since I finished my first degree!

Alright cool cheers.

In my day we had to work this sh*t out ourselves 😆

alright grandpa.. 😉

i get dy/dx as (3x^2)(e^3x)+(2ex^3x)

I think GJP is right. So are you. Assuming you mean 2xe^3x rather than 2ex^3x in the second bit.

Are you doing AS level maths? I am learning differentiation at the moment

no, just finished as, on to a2 now. also doing as further and a2 further (all in one year..)

good luck 😉
your working looks right, but since ive only just learnt this i couldnt be sure

Realman - I wasn't trying to confuse you or lead you to believe your calculations were incorrect. I found it easier to work through it myself and only at the end did I realise we had done the same thing.

I am confident this is the right answer - the only thing I wasn't 100% sure on was the first derivative of e^3x ( I was a seriously sh1t hot mathematician in my day - if there is such a thing!)

Good luck with the studies.
Gary

All i know is soon as you leave school/college/uni it will be the last time you ever need this!

Used to get stressed about this myself but never needed it since!

🙁 😕 🙂 😀 8) 😆 😉

I spent years and years learning this stuff and I can honestly say that none of it has ever been of any use to me whatsoever. I always hated maths anyway.
Then again if you get an engineering degree and then don't become an engineer I suppose that shouldn't be so much of a surprise.

I believe the answer to be y=e^3(5x-4)

Edit: Check your original post, you have the constant c as -4e^3 yet in your final answer you have it as -4x^3

I'm sure this is a typo and what you meant is the above answer.

A tip though: Don't use y=mx+c any more, use y-y1=m(x-x1)
Although using y=mx+c is ok for these questions once you start doing harder ones especially if you are doing further maths ( Fp1, Fp2 and Fp3) it makes it easier for you.

Once painfully witnessed two girl shop assistants trying to work out 15% of £1. Took them about 5 minutes, and they still had their doubts.