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Imaan49

Need help in Calculas

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:D

 

i cant get this stuff through my head, does anybody know anything about derivatives

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salaam alaikum,

 

i know derivatives. I would rank my knowledge of them at an 8 of 10. 10 being math teacher knowledge.

 

feel free to ask questions.

 

wassalaam,

AS

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ok umm i hope u try to understand this question then

 

Find the derivative for each of the following functions.Express you answer in a fully simplified factored form:

 

y=(4-x^2)^2 multiplied by the cubed root of (5x^3 -8)^2

 

i know this looks confusing, i cant find some of the buttons on my keyboard so i'll just tell u, 4 minus x squared multiplied by the cubed root of 5 x cubed minus 8 all squared :D i hope this makes sense

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selam,

 

editted....

Edited by elif74

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asslamualaikum

 

hey sis have u done this dy/dx??

 

do you use that for differentiation?

 

I am studying that now, if that is it I can give u a hand, by the way the first one is it (4-x^2) all squared??

 

walaikumsalam

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no its not all squared, just the x is squared, an ya it is dy/dx..if u want to do it like that, i dont even remember the formula for the product rule

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asslamualaikum

 

ok sister, i will try it, but can you also clarify this for me

 

multiplied by the cubed root of 5 x cubed minus 8 all squared

 

is that the cube root of (5x^3 - 8) all squared??

 

walaikumsalam

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:D

 

yes it is

 

it would b easier if u saw my worksheet

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salaam alaikum,

 

so it's:

 

y=(4-x^2)^2 * ((5x^3-8)^2)^1/3 ?before differentiation?

 

 

wassalaam,

AS

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:D

 

yes lool, y didnt i just write it like that,

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salaam alaikum,

 

I haven't done it by hand yet, but running it numerically gives me:

 

4x*(x^2-4)*(5x^3-8)^(2/3)+(10*x^2*(x^2-4)^2)/(5x^3-8)^(1/3)

 

I hope that helps. If it needs to be shown step by step i will have to do that later, i posted this for the sake of giving you an answer.

 

wassalaam,

AS

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:D

 

lool ur too kind bro..but if u dont mind i would like to know the other step :D

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selam,

 

sorry for my log'ic approach that i editted...i mean too many years didnt look at calculus book after some study, i think i found it....

 

firstly the rule of the multiplication is:

 

if y = z.t (dot is multiply *)

 

y' (that is derivative you may write [f'(x) or dy/dx] dx shows the derivative with respect to x)

 

y' = z'.t + z.t'

 

and your example:

 

y=(4-x^2)^2 * ((5x^3-8)^2)^1/3

 

if z = (4-x^2)^2

 

you may think that a = (4-x^2)

 

the equation will become:

 

z = a^2

 

dz / da = 2a. a'

 

z' (or dz/dx) = 2.(4-x^2).(-2x)= (-4x)(4-x^2)= 4x (x^2-4)

 

and the other part

 

t= ((5x^3-8)^2/3

 

if b= (5x^3-8)

 

t = 3?b

 

dt/db = 2/3. b^(-1/3). b'

 

b' = db/dx= 15 x^2

 

and dt/db = 2/3. (5x^3-8)^(-1/3). 15x^2

 

and the last part the equation:

 

y=(4-x^2)^2 * (5x^3-8)^2/3

 

dy/dx = (4x)(x^2-4)[(5x^3-8)]^2/3 + [(4-x^2)^2].2/3. (5x^3-8)^(-1/3). (15x^2)

 

dy/dx = (4x)(x^2-4)[(5x^3-8)]^2/3 + (10x^2)[(4-x^2)^2].(5x^3-8)^(-1/3)

 

you know [(4-x^2)^2] = [(x^2-4)^2]

 

and (5x^3-8)^(-1/3) = 1 / (5x^3-8)^(1/3)

 

this is the same solution with Allah's slave

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salaam alaikum,

 

Imaan49: you have your answer and im glad we could help, i hope it is not too late for your homework deadline or something. :D

 

elif74: thank you for taking the time to break down the solution into steps, as well as verifying my answer. :D

 

w/s,

AS

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:D

 

thank u sooo much, all of u guys, im soo happy now :D

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lol i'm happy you got your question solved!

 

 

 

 

 

 

 

 

:D but wait till you get into partial differentiation and 3D derivatives LOL (w00t)!!

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:D

 

ur freaking me out dh

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^^ bro (or sis (w00t)) no worries..

 

cause those neat things will not get you unless you enroll yourself in engineering or some other post secondary program with a lot of math.. hehe.. B)

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selam,

 

 

cs, you cant write like this but

 

 

d(4x^2) /dx = 8x

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:D

 

im a sis, doesnt my screen name give that away :D

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:D

What year are you in?

 

I am in year 12 and i got exam 2 days after about all book, and i also hate this derivative things, the question u wrote i can do that, but its long way...

 

Their are easy to learn , using chain rule will solve the first part and second, then use product rule to join it together....

Get a book of year 12 maths and you will be fine , its chapter 7 in my book, hhaaha..

Derivative is fine, but antiderivative is also fine, its just a mater of understanding..

 

example

Find derivative of

2x^2 = 4x

 

find antiderivative of 4x

4x= 4x^2/2

= 2x^2

these are easy, the dam one is log one and exponential... they are also easy haha..

 

i got problem with how to get exact value...

i mean how do u get exact value of 2.07, like in form of square root..

 

Good luck... "He teaches man what he knows not" Quran..

Pray to Allah and ask Him to increase your knowlege, you will be fine..

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lol anyone know the derivative of:

 

e^ln(x) :D ??

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selam,

 

haha this became really hot topic....

 

dh asking really or jokes??? :D

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