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#1
can someone plz help me on this differentiation question.

differentiate y= sqrt(3x^2 +2)

I got: 1/2(3x^2 +2)^(-1/2)

is this right?
I don't know what is going on in the marking scheme. Q10.
http://www.dunblanehighschool.org.uk...lutions-SA.pdf

thanks.
also, I am gonna try n get a lot of maths done today as I have a prelim tomorrow.
so, plz lend a hand if I am stuck

thanks again.
0
1 year ago
#2
You need to use chain rule to differentiate this.
If you are not comfortable doing it directly, as done in the mark scheme, you should write it out explicitly:
y = sqrt(3x^2 - 2) becomes
y = sqrt(u) and u = 3x^2 - 2
dy/dx = dy/du x du/dx
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#3
(Original post by JSG29)
You need to use the chain rule to differentiate this.
If you are not comfortable doing it directly, as done in the mark scheme, you should write it out explicitly:
y = sqrt(3x^2 - 2) becomes
y = sqrt(u) and u = 3x^2 - 2
dy/dx = dy/du x du/dx
I am pretty sure I have not learnt the chain rule. can u plz elaborate on how to do it. thanks a lot.
0
1 year ago
#4
(Original post by dude101010)
can someone plz help me on this differentiation question.

differentiate y= sqrt(3x^2 +2)

I got: 1/2(3x^2 +2)^(-1/2)

is this right?
I don't know what is going on in the marking scheme. Q10.
http://www.dunblanehighschool.org.uk...lutions-SA.pdf

thanks.
also, I am gonna try n get a lot of maths done today as I have a prelim tomorrow.
so, plz lend a hand if I am stuck

thanks again.
Is this Y13 or Y12 Maths?
0
#5
(Original post by Jackudy3)
Is this Y13 or Y12 Maths?
I am in Scotland lol.
I am doing Scottish Highers. they are in between GCSE's and A-levels ( i think).
0
1 year ago
#6
When you have composite functions (functions of functions) like in the example, you need to differentiate the function in separate parts.
First, differentiate the outer function, taking what is inside as a constant term. Then differentiate the inner function, and multiply the 2 derivatives together. The example should make this make more sense

In the example, you have the square root of a polynomial, so you set
u = 3x^2 - 2
Then y = sqrt(u)
To differentiate y, you differentiate y with respect to u (dy/du) and multiply it by du/dx
So dy/dx = (dy/du)(du/dx)
dy/du = 1/(2sqrt(u)) , du/dx = 6x

So dy/dx = 6x/(2sqrt(u)) = 3x/sqrt(3x^2-2)

When doing this, it is important to remember to substitute 3x^2 - 2 back for u
1
#7
(Original post by JSG29)
When you have composite functions (functions of functions) like in the example, you need to differentiate the function in separate parts.
First, differentiate the outer function, taking what is inside as a constant term. Then differentiate the inner function, and multiply the 2 derivatives together. The example should make this make more sense

In the example, you have the square root of a polynomial, so you set
u = 3x^2 - 2
Then y = sqrt(u)
To differentiate y, you differentiate y with respect to u (dy/du) and multiply it by du/dx
So dy/dx = (dy/du)(du/dx)
dy/du = 1/(2sqrt(u)) , du/dx = 6x

So dy/dx = 6x/(2sqrt(u)) = 3x/sqrt(3x^2-2)

When doing this, it is important to remember to substitute 3x^2 - 2 back for u
thanks a lot, mate. this is some advanced stuff! i kinda understand it but what did you do to get 6x?

thanks again for ur time
0
1 year ago
#8
we have u = 3x^2 - 2
so du/dx = d/dx (3x^2 - 2) = 2*3x = 6x
0
#9
cheers.
I didn't know you can actually substitute into e.g. du/dx.
0
#10
how do u turn this

x + 4/x =b

into ax^2 +bx + c form.
0
1 year ago
#11
Multiply both sides of the equation by x

x(x+4/x) = bx
x^2+4=bx
0
#12
(Original post by JSG29)
Multiply both sides of the equation by x

x(x+4/x) = bx
x^2+4=bx
you're a legend
0
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