# Core 3 DifferentiationWatch

#1
Differentiate the following with respect to x.

x^3(x+1)^5

I understand you use product rule, and for the first bit I get...
3x^2(x+1)^5

But I don't understand how to do the second part ://
0
9 years ago
#2
its the chain rule within the product rule
0
9 years ago
#3
(x^3) * (x+1)^5

u=x^3 du/dx=3x^2
v=(x+1)^5 dv/dx=5(x+1)^4. Use the chain rule to differentiate (x+1)^5.

chain rule: y = (x+1)^5 = z^5 when z=x+1.
dy/dz = 5x^4 dz/du = 1
so dy/dx= dy/dz * dz/du = 5x^4

You need to know this chain rule for basically all integrations and differentiations.
0
9 years ago
#4
First one multiply differential of second PLUS second one multiple differential of the first.
0
9 years ago
#5
You need to use the chain rule to differentiate

to do this, let

then,

chain rule:

so

and

so
0
9 years ago
#6
Is the question x^(3(x+1)^5) or (x^3)*((x+1)^5)?
0
#7
Yeah I got the same as you guys, but the OCR Mark scheme has a different answer...

http://www.ocr.org.uk/Data/publicati..._MS_Jun_07.pdf

Go to page 16, Question 1.
0
9 years ago
#8
u mind typing the answer out, the attachment isn't opening from my comp right now
0
9 years ago
#9
I got the same answer as them:

dy/dx of x^3 is 3x^2 so there you get 3x^2(x+1)^5
for the second part of the product rule its using the chain rule of (x+1)^5 which is 5(x+1)^4 x x^3

therefore you get 3x^2(x+1)^5 + 5x^3(x+1)^4
0
#10
(Original post by CCJ)
I got the same answer as them:

dy/dx of x^3 is 3x^2 so there you get 3x^2(x+1)^5
for the second part of the product rule its using the chain rule of (x+1)^5 which is 5(x+1)^4 x x^3

therefore you get 3x^2(x+1)^5 + 5x^3(x+1)^4
Where do you get the 'TIMES x^3 from though?
0
9 years ago
#11
Where do you get the 'TIMES x^3 from though?
product rule = v du/dx + u dv/dx

you use the chain rule to get dv/dx being 5(x+1)^4 x x^3, and since u is x^3 therefore you multiply them to get 5x^3(x+1)^4 by definition of the product rule.
0
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