Differentiate (1 Viewer)

[boongsta]

hey guys test tomorrow one thing i need help with ....


Differentiate with respect to x

3x-5/2x+1

please show all working i really need to see it

thanks heaps

 

[yrtherenonames]

Quotient rule my man
y=3x-5/2x+1
Let v=2x+1 Let u=3x-5
dv/dx=2 du/dx=3
dy/dx=(vu'-uv')/v^2
=3(2x+1)-2(3x-5)/(2x+1)^2
=13/(2x+1)^2

 

[Dreamerish*~]

hey guys test tomorrow one thing i need help with ....


Differentiate with respect to x

3x-5/2x+1

please show all working i really need to see it

thanks heaps

3x-5/2x+1

d/dx = (vu' - uv')/v²
= [(2x+1)3-(3x-5)2]/(2x+1)²
= (6x+3-6x+10)/(2x+1)²
= 13/(2x+1)²

 

[Jago]

on a side note, please use brackets.

 

[Dreamerish*~]

Thanks a lot makes perfect sense except im not sure what u mean by (vu'-uv') ... what do the ' things mean?

y' = dy/dx = f'(x)

(In explicit equations)

 

[insert-username]

The ' simply means the derivative of u and the derivative of v. vu' = v x the derivative of u.

Hmmm... funky thread errors all over again...


I_F

 

[boongsta]

Thanks a lot makes perfect sense except im not sure what u mean by (vu'-uv') ... what do the ' things mean?

 

[insert-username]

Random board errors. They've been popping up a lot today - don't worry too much about it.


I_F

 

[boongsta]

thanks a lot guys helps heaps .......... except why the hell is my post last post below the replies to that post lol

 

[insert-username]

Alright. Basically:

if y = u ÷ v

dy/dx = (vu' - uv') ÷ v²

Every time you do a question using that rule, write it out. You'll remember it once you've written ten times or so. The numerator is u and the denominator is v. The best way to remember it isn't by sitting there looking at it - do questions using the rule and you'll learn it much quicker.


I_F

 

[boongsta]

have u guys got any ideas to remember this easily ... or maybe someone can explain the reasoning behind the solution? its kinda hard to memorise this

 

[Dreamerish*~]

have u guys got any ideas to remember this easily ... or maybe someone can explain the reasoning behind the solution? its kinda hard to memorise this

You need to remember these rules:

Product rule: uv' + vu'

Example: differentiate (3x + 4)(5x + 5)
u = 3x + 4
u' = 3
v = 5x + 5
v' = 5
uv' + vu' = (3x + 4).5 + (5x + 5).3
= 15x + 20 + 15x + 15
= 30x + 35

Quotient rule: (vu' - uv')/v²

Example: differentiate (2x + 5)/3x
u = 2x + 5
u' = 2
v = 3x
v' = 3
(vu' - uv')/v² = [3x.2 - (2x + 5).3]/9x²
= (6x - 6x - 15)/9x²
= -15/9x²

Chain rule: Not sure how I can put this, so just look at the example.

Example: differentiate (4x + 16)¹⁵
Look at (4x + 16) as one thing, and differentiate. You get:
15(4x + 16)¹⁴. However, you need differentiate the "insides" (that is, 4x + 16) and multiply it by the 15(4x + 16)¹⁴.

So we end up with: 4.15(4x + 16)¹⁴
= 60(4x + 16)¹⁴

If anyone has a better explanation of the chain rule, post it up.

 

[boongsta]

thanks HEAPS for the help

im just gunna type out the formula here to see if i get it right lol

(3x-5)/(2x+1)

numerator = u
denominator = v

du/dx = differentiated numerator
dv/dx = differentiated denominator

dy/dx = (vu'-uv')/v^2

is that right mate - e - o - s?

does it matter whether u and v and numerator and denominator or can it be the other way around cause form the formula it looks like it doesn't matter

 

[Templar]

Chain rule...dy/dx = dy/du * du/dx

Or, f(x)=g(h(x))
f'(x)=g'(h(x))*h'(x)

 

[rnitya_25]

Chain rule...dy/dx = dy/du * du/dx

Or, f(x)=g(h(x))
f'(x)=g'(h(x))*h'(x)

that just complicates things so much its not funny. sorry, but i had to say. what dreamerish said is all you need to know, in perfect format.

 

[insert-username]

You'll apparently need the chain rule later on, so it's probably worth learning now. It's basically the function of a function rule - where u is the function without the exponent.

dy/dx = n.f'(x).f(x)^n-1

dy/dx = dy/du * du/dx

It's confusing on the screen, but written out it makes much more sense.


I_F

 

[Jago]

1st times the derivative of the 2nd + the 2nd times the derivative of the 1st.

i've never learnt that v u crap...:/

Edit: come to think of it, that may be why i had so much trouble with rates of change in 3u hahahah

 

[boongsta]

lol thanks a lot fellas im bookmarking this page in case of further reference ... maybe u should copy paste wat u said dream into a document and submit it 2 bos.org?

thanks again

 

[pLuvia]

hey guys test tomorrow one thing i need help with ....


Differentiate with respect to x

3x-5/2x+1

please show all working i really need to see it

thanks heaps

'

y = 3x-5 / 2x+1
y' = [(2x+1)(3) - (3x-5)(2)] / (2x+1)^2
y' = [6x+3 - 6x+10] / (2x+1)^2
y' = 13 / (2x+1)^2

 

[haboozin]

hey guys test tomorrow one thing i need help with ....


Differentiate with respect to x

3x-5/2x+1

please show all working i really need to see it

thanks heaps

dont need any losery rule....


If you dont want to learn those rules here is an alternative method
(3x - 5)/(2x + 1) = 3/2 - 13/2(2x + 1)


=3/2 -13/2(2x + 1)^-1
dy/dx= 13/(2x + 1)^2


OR

by first principles

lim h-->0

(f(x + h) - f(x))/h