Mei C3 differentiation helpWatch

#1
How do i differentiate y=x-2/(x+3)^2 and y=(x+1) sqrt(x-1)

The first i've tried using the chain rule then the quotient rule but cannot get the right answer can anyone help and show the working
0
4 years ago
#2
(Original post by James2015)
How do i differentiate y=x-2/(x+3)^2 and y=(x+1) sqrt(x-1)

The first i've tried using the chain rule then the quotient rule but cannot get the right answer can anyone help and show the working
Ambiguous notation on the first.

Product rule on the second.
0
#3
(Original post by Mr M)
Ambiguous notation on the first.

Product rule on the second.
First one is (x-2)/(x+3)^2
0
4 years ago
#4
(Original post by James2015)
First one is (x-2)/(x+3)^2
Product rule on both then.

You need to know that
0
4 years ago
#5
(Original post by James2015)
First one is (x-2)/(x+3)^2
Use the quotient rule.

Diferentiate (x-2) normally.
Differentiate (X+3)^2 using the chain rule

Then use the quotient rule formula, and factorise etc if applicable.
1
4 years ago
#6
(Original post by Rickstahhh)
Use the quotient rule.

Diferentiate (x-2) normally.
Differentiate (X+3)^2 using the chain rule

Then use the quotient rule formula, and factorise etc if applicable.
Unnecessarily complicated.
0
4 years ago
#7
(Original post by Mr M)
Unnecessarily complicated.
You can't use product rule on this, as there is not two functions of x multiplied together.

Imo quotient rule is the easiest option
0
4 years ago
#8
(Original post by Rickstahhh)
You can't use product rule on this, as there is not two functions of x multiplied together.
Are you sure?

0
4 years ago
#9
(Original post by Rickstahhh)
You can't use product rule on this, as there is not two functions of x multiplied together.

Imo quotient rule is the easiest option
Yes there are: and .
0
#10
(Original post by Mr M)
Unnecessarily complicated.
I'm still confused with the first equation.

u=(x-2)
u'=1

v=(x+3)^2
v' = 2(x+3)

Sub these into (v*u' - u*v')/(v^2)

Is that right?
0
4 years ago
#11
(Original post by James2015)
I'm still confused with the first equation.

u=(x-2)
u'=1

v=(x+3)^2
v' = 2(x+3)

Sub these into (v*u' - u*v')/(v^2)

Is that right?
You can do that but it is long winded. Your work so far is correct.
0
4 years ago
#12
(Original post by Mr M)
Are you sure?

Yeah, for the question (x-2)/(x+3)^2
As it is a rational function in the form g(x)/h(x), where g(x) and h(x) are functions.

Because , which differentiates to
0
4 years ago
#13
(Original post by Rickstahhh)
Yeah, for the question (x-2)/(x+3)^2
As it is a rational function in the form g(x)/h(x), where g(x) and h(x) are functions.

Because , which differentiates to
No way that could possibly be written as a product then (ignoring Smaug's amusing contribution)?
0
4 years ago
#14
(Original post by Rickstahhh)
You can't use product rule on this, as there is not two functions of x multiplied together.

Imo quotient rule is the easiest option
You can. Think about negative exponents.

(Original post by Smaug123)
Yes there are: and .
Technically correct. Also totally useless.
0
4 years ago
#15
(Original post by james22)
Technically correct. Also totally useless.
<doffs cap>
1
4 years ago
#16
use chain and product together doy
0
4 years ago
#17
(Original post by Mr M)
No way that could possibly be written as a product then (ignoring Smaug's amusing contribution)?
Or you could do (x-2)(x+3)^-2
Then use product rule?
0
4 years ago
#18
(Original post by Rickstahhh)
Or you could do (x-2)(x+3)^-2
Then use product rule?
That's what I would do. It is usually best to avoid the quotient rule if the denominator is simple.
0
#19
First equation [(x+3)^2 - (x-2)*[2(x+3)]]/(x+3)^4

has to factor down to (7-x)/(x+3)^3
0
4 years ago
#20
(Original post by Mr M)
That's what I would do. It is usually best to avoid the quotient rule if the denominator is simple.
Yeah I totally understand what you mean. I just worked it out using both the quotient and product rule, and tbh the product rule is much much faster
1
X

new posts
Latest
My Feed

Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

see more

See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

University open days

• Cranfield University
Cranfield Forensic MSc Programme Open Day Postgraduate
Thu, 25 Apr '19
• University of the Arts London
Open day: MA Footwear and MA Fashion Artefact Postgraduate
Thu, 25 Apr '19
• Cardiff Metropolitan University
Sat, 27 Apr '19

Poll

Join the discussion

Have you registered to vote?

Yes! (293)
37.61%
No - but I will (56)
7.19%
No - I don't want to (56)
7.19%
No - I can't vote (<18, not in UK, etc) (374)
48.01%