If y = (2x - 5)⁷ , find dy/dx

happy sunday my people.........hope say una never forget d latest news oh......jamb has decided to use pepper to rub people's eyes next yr by anoder method ooo

Halpher:
Yea relating to its High proportion of Oxygen.

d question shudn't have dat option.

urmmm.
avin problems wt physics.... nid lil help...
+ where's everybdy?

mathefaro: If y = (2x - 5)⁷ , find dy/dx

D ansa is 14(2x-5)[sup][/sup]6

mathefaro: If y = (2x - 5)⁷ , find dy/dx

14(2x-5)⁶

chimhigher: 14(2x-5)⁶

Correct, but you should have shown the house how you got the answer na.

What is the length of an arc which is subtended by an angle of 60⁰ from the circumference of a circle of radius 9cm in terms of pi

mathefaro: Correct, but you should have shown the house how you got the answer na.

owk m sowi rum..2x-5)⁷ .............Let (2x-5) be U...so it becomes U⁷ so dy/du=7U⁶ dfferentiating (2x-5) dy/dx=2. Xo dy/du X du/dx= 2*7U⁶= 14(2x-5)⁶..
Dax it oga boss

mathefaro: What is the length of an arc which is subtended by an angle of 60⁰ from the circumference of a circle of radius 9cm in terms of pi

lenght f an arc= @/360*2πr.......so 60/360*2π*9=1/4*2π*9=18π/4= 4.5π

chimhigher: lenght f an arc= @/360*2πr.......so 60/360*2π*9=1/4*2π*9=18π/4= 4.5π

I know you'll fall for the trap, this answer is wrong, go through the question again please, I know you'll get it.

mathefaro: I know you'll fall for the trap, this answer is wrong, go through the question again please, I know you'll get it.

lol sowi jawe 2 much rush rush...i didnt c d circumference i tut twas centre...k.angle at d circumference= 2 X angle n d Centre..so its 120....so 120/360*2π9... daz 1/3* 2π9= 18π/3=6π. Lol rush rush no gud...

chimhigher: lol sowi jawe 2 much rush rush...i didnt c d circumference i tut twas centre...k.angle at d circumference= 2 X angle n d Centre..so its 120....so 120/360*2π9... daz 1/3* 2π9= 18π/3=6π. Lol rush rush no gud...

exactly

Given that dy/dx = x + 1/x, find the value of y.

mathefaro: Given that dy/dx = x + 1/x, find the value of y.

d ansa is 1-x^-2 or 1-1/x²

chimhigher: d ansa is 1-x^-2 or 1-1/x²

I'm sorry, this is wrong bro. Go over the question again

X+1/x wuld hv used d first principle buh i cant typ it...so lemme jes go on ....dy/dx of (X)=1 nd 4 1/x from indices its X^-1 xo using differentiation principle nx^n-1 xo it bcomes -1x^-1-1 i.e -1x^-2 so d ansa is 1+(-1x^-2= 1-1/x²

Lol ooooh sowi 1ncemur..ummm d ansa wuld b 1/2x²- 1/2x⁰ m i ryt?? Kinda confusing

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chimhigher: Lol ooooh sowi 1ncemur..ummm d ansa wuld b 1/2x²- 1/2x⁰ m i ryt?? Kinda confusing

calm down, you still haven't gotten what the question is all about, and that's why you're getting it all wrong.

Hint : integration

chimhigher: X+1/x wuld hv used d first principle buh i cant typ it...so lemme jes go on ....dy/dx of (X)=1 nd 4 1/x from indices its X^-1 xo using differentiation principle nx^n-1 xo it bcomes -1x^-1-1 i.e -1x^-2 so d ansa is 1+(-1x^-2= 1-1/x²

PrinceCEE: u were correct by saying that 'I' is singular.


Point of correction brother::::::::> Has and Have are present tense used for singular and plural nouns respectively while Had is the past tense of both.



Hope I ve cleared u

you are correct

update resume tonight

mathefaro: Given that dy/dx = x + 1/x, find the value of y.

First and foremost, we need to understand that integration is anti-derivation/ anti-differentiation I.e the opposite of differentiation. If y = x², dy/dx (differentiation of y with respect to x) = 2x.
Thus, since dy/dx = 2x, y = x²(integration) [ hint: the integral of axⁿ = (ax^{n + 1})/ (n + 1)

Now to the question, dy/dx = x + 1/x (which means that the differentiation of yywith respect to x is x + 1/x)
To get y, we need to integrate x +1/x;
= x²/2 + lnx + k

NB:
* using the formula given above, the integration of x =x²/2
*the integration of 1/x = ln x ( logarithm of x to base e) since the differentiation of lnx = 1/x
* k is constant which must be included in every integral solution

how una dey my people

Benjaminbest: how una dey my people

Gud...what of the update??
following....

after working today i discovered dat i cannot update again because d stress is tooooooooo much. Tomoro i will update. Abeg make una no vex body nor b fire wood.....hope una understand

igstars:

thankz for creating dz thread nd kipping it u....
am in support of d whatsapp group chat idea, @least dat way i cn get notified of updates nd I for one wud try 2 contribute as a learner sha.....
plsss der'll b mre progress dat way!
nd u knw as tinz r nt evry1 visits d net evryday.....

ohhh
whizzkid u're here....
wat abt prof123?

No ma...By God's grace, I've been admitted already....Someone shared d link in another thread mentioning that someone in the shared link's thread needed an explanation on a chemistry concept..I entered so dat I could contribute and also learn something new...I think d matter has been cleared now anyway..
I can't remember u now, but m sure u were once among the members of the post utme thread created by prof1234...I haven't seen d guy's post on NL for a long time now....So how're u doing?

pls help me to solve this...its physics question....luk simply...but kind of complicated.

An aircraft flies a distance of 5km from a town A, on a bearing of 030 degree to a town B. The bearing of A from B would be?

thank you

happy weekend pals....i must update today first i nid questions dat will ginger me up. So guys kip it coming

Destilo:
pls help me to solve this...its physics question....luk simply...but kind of complicated.

An aircraft flies a distance of 5km from a town A, on a bearing of 030 degree to a town B. The bearing of A from B would be?

thank you

It's a mathematics question
If the bearing B from A is 030, then the bearing of A from B is 210^o or S30^oW.