The Student Room Group
Uni students - Complete our survey >>
Uni students - Complete our survey >>
Uni students - Complete our survey >>

differentiation

Please could someone help with this question.

Given that x^3+y^3-3x-6=0 find dy/dx in terms of x and y.Hence find the turning points.

dy/dx= 3-3x^2/3y^2

How do i find the turning points?
dy/dx = 0 at turning points
Reply 2
Avatar for roseanne_12
roseanne_12
OP
is it not (3-3x^2) all divided by 3y^2 ?
Reply 3
Avatar for roseanne_12
roseanne_12
OP
Original post by Felix Felicis
dy/dx = 0 at turning points

Thank you :smile:

I know that it when dy/dx=0
however I'm stuck because i got (3-3x^2)/3y^2=0
I don't know how to solve it from there :/
Reply 4
Avatar for ben494
ben494
Original post by roseanne_12
Thank you :smile:

I know that it when dy/dx=0
however I'm stuck because i got (3-3x^2)/3y^2=0
I don't know how to solve it from there :/


Top line must be equal to 0.

So 3-3x^2=0
So 3x^2=3
So x^2=1
X=+or-1
Then solve for y


Posted from TSR Mobile
Reply 5
Avatar for roseanne_12
roseanne_12
OP
Original post by ben494
Top line must be equal to 0.

So 3-3x^2=0
So 3x^2=3
So x^2=1
X=+or-1
Then solve for y


Posted from TSR Mobile


Thank you very much! you really helped!

Could you help me with something else please.It's probably quite straightforward but i dont really understand.
What does it mean when it says differentiate with respect to x e.g.y^3
do i write it out as dx/dy ?
Reply 6
Avatar for Goods
Goods
16
use implicit differentiation

x^3+y^3-3x-6=0

goes to

3x^2+3y^2 dy/dx -3 = 0

therefore dy/dx = 1-x^2/y^2

to make the fraction = 0 x either has to be 1 or -1 to make the numerator 0 denominator doesn't matter
Reply 7
Avatar for Goods
Goods
16
Original post by roseanne_12
Thank you very much! you really helped!

Could you help me with something else please.It's probably quite straightforward but i dont really understand.
What does it mean when it says differentiate with respect to x e.g.y^3
do i write it out as dx/dy ?


differentiate it with the chain rule, so differentated (y)^3 then multiply it by the derivative of the bracket which is dy/dx. respect to x means dx is the denominator.
Reply 8
Avatar for roseanne_12
roseanne_12
OP
Original post by Goods
differentiate it with the chain rule, so differentated (y)^3 then multiply it by the derivative of the bracket which is dy/dx. respect to x means dx is the denominator.


Great help! thank you so much :smile:
Reply 9
Avatar for TenOfThem
TenOfThem
18
Original post by roseanne_12
Thank you :smile:

I know that it when dy/dx=0
however I'm stuck because i got (3-3x^2)/3y^2=0
I don't know how to solve it from there :/


For this type of question there is no need to write dy/dx as a fraction

You can simply substitute dy/dx=0 into the implicitly differentiated equation

This is worth realising, especially when you have more complicated equations with more than one function of y

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you