You are Here: Home >< Maths

# Can someone check if this C3 dy/dx ques is right? watch

1. find dy/dx of

y = (x+2)^2 / (x-1)^4

(-x^2 - 4x + 4) / (x-1)^5

Thanks
2. (Original post by kiiten)
find dy/dx of

y = (x+2)^2 / (x-1)^4

(-x^2 - 4x + 4) / (x-1)^5

Thanks
Maple disagrees so unfortunately you got it wrong

3. (Original post by kiiten)
find dy/dx of

y = (x+2)^2 / (x-1)^4

(-x^2 - 4x + 4) / (x-1)^5

Thanks
4. (Original post by kiiten)
find dy/dx of

y = (x+2)^2 / (x-1)^4

(-x^2 - 4x + 4) / (x-1)^5

Thanks

http://www.wolframalpha.com/input/?i...+%2F+(x-1)%5E4
5. Was i right to use the quotient rule where:

(u^1v - uv^1 ) / v^2 ?
6. (Original post by B_9710)
Maybe im missing something here or just not seeing it at all but how did they get to that last step here?
7. (Original post by kiiten)
Was i right to use the quotient rule where:

(u^1v - uv^1 ) / v^2 ?
Quotient or product, doesn't really matter which as long as your working is consistent. Show your working if you want your error to be pointed out.

(Original post by kiiten)
Maybe im missing something here or just not seeing it at all but how did they get to that last step here?
They put the first term under the same denominator as the second by multiplying top and bottom by (x-1) then just factorised the numerator fully.
8. (Original post by kiiten)
Maybe im missing something here or just not seeing it at all but how did they get to that last step here?
They made both fractions have a common denominator, then probably expanded the brackets and simplified.
9. (Original post by RDKGames)
Quotient or product, doesn't really matter which as long as your working is consistent. Show your working if you want your error to be pointed out.

They put the first term under the same denominator as the second by multiplying top and bottom by (x-1) then just factorised the numerator fully.
(Original post by Jasaron)
They made both fractions have a common denominator, then probably expanded the brackets and simplified.
If i do that i get this. Ill attach my working (i kinda scribbled some of it out xD)

NOTE: I made a = x+2 and b= x-1

Posted from TSR Mobile
Attached Images

10. (Original post by kiiten)

Posted from TSR Mobile
Expand the thing you get at the end, (2ab-a^2 . b) and re-factorise, taking the -2 out of the brackets and you'll (probably) get the answer that the computer is giving you.
11. (Original post by kiiten)
If i do that i get this. Ill attach my working (i kinda scribbled some of it out xD)

NOTE: I made a = x+2 and b= x-1

Posted from TSR Mobile
That's right. Now you can factorise the numerator fully.
12. (Original post by RDKGames)
That's right. Now you can factorise the numerator fully.
Yeah I got 2(x+2)(x-1) - 4(x+2)(x+2) which expands to the answer they have.

Thanks everyone
13. (Original post by RDKGames)
That's right. Now you can factorise the numerator fully.
If you derive e^x(x+2)^2

the calculator says you get (x+4)(x+2)e^x
Can you divide the whole thing by e^x to leave x^2 + 6x + 8

If so would the stationary points be -4, 4e^x and -2, e^x
14. (Original post by kiiten)
If you derive e^x(x+2)^2

the calculator says you get (x+4)(x+2)e^x
Can you divide the whole thing by e^x to leave x^2 + 6x + 8

If so would the stationary points be -4, 4e^x and -2, e^x
Er... Well if you JUST derive it, then no, you cannot divide an expression - only factor it.

If you're finding the stationary points and make it equal to 0, then loosely speaking you can indeed just divide by but strictly speaking you need a note as to why you're allowed to take this step.
15. (Original post by RDKGames)
Er... Well if you JUST derive it, then no, you cannot divide an expression - only factor it.

If you're finding the stationary points and make it equal to 0, then loosely speaking you can indeed just divide by but strictly speaking you need a note as to why you're allowed to take this step.
So if the expression is equal to something then you can divide by a factor?

Does that mean the stationary points could be:
-4, 4e^x
-2, e^x
e^x, ?
16. (Original post by kiiten)
So if the expression is equal to something then you can divide by a factor?
Well yeah. These are called equations so you can divide both sides by whatever you want as long as it's not 0.

It wouldn't make sense to divide on its own by x because then you get 1 - and this is a different quantity than what x is. If you had then you are allowed to divide by x on both sides if x does not equal 0.

Does that mean the stationary points could be:
-4, 4e^x
-2, e^x
e^x, ?
First two yeah, but not the last one because is not a solution to the differentiated equation.
17. (Original post by RDKGames)
Well yeah. These are called equations so you can divide both sides by whatever you want as long as it's not 0.

It wouldn't make sense to divide on its own by x because then you get 1 - and this is a different quantity than what x is. If you had then you are allowed to divide by x on both sides if x does not equal 0.

First two yeah, but not the last one because is not a solution to the differentiated equation.
The differentiated equation is e^x(x+4)(x+2) which i made =0 to find the stationary points. Is e^x not a solution because it doesnt have an x in a bracket like the other 2? (sorry that was badly worded)

NOTE: I have a bunch of other questions that ive answered but need checking. If i post them here would you be able to check them?
18. (Original post by kiiten)
The differentiated equation is e^x(x+4)(x+2) which i made =0 to find the stationary points. Is e^x not a solution because it doesnt have an x in a bracket like the other 2? (sorry that was badly worded)
No. Try solving

NOTE: I have a bunch of other questions that ive answered but need checking. If i post them here would you be able to check them?
Go for it.
19. (Original post by RDKGames)
No. Try solving

Go for it.
lne^x = ln0
x= ? (invalid).

Thanks

1. Find coordinates of y = e^x when the gradient is 3
is x = ln3 and y =3

2. Find equation of tangent to y= e^x at (3, e^3)
y - e^3 = e^3(x - 3)
y= xe^3 - 2e^3

3. Equation of normal to y=lnx at (3, ln3)
y - ln3 = -x(x - 3)
y = -x^2 + 3x + ln3

4. find nature of the stationary point x=ln2e dy/dx=2e-e^x
d2y/dx2 = -e^x
d2y/dx2 = -e^ln2e
d2y/dx2 = -2e therefore max. point

5. Estimate gradient of curve y = (x+lnx)^2 when x=3
change in y / change in x
(0.0601 + ln0.0601) / 0.01

- let me know if anything is unclear or if you want me to take a picture of my actual working. Thanks again
20. (Original post by kiiten)
lne^x = ln0
x= ? (invalid).

Thanks
Precisely, there is no solution to that. This can easily be seen from remembering the graph of and how it is strictly positive for all values of x (and by strictly positive it means )

1. Find coordinates of y = e^x when the gradient is 3
is x = ln3 and y =3

2. Find equation of tangent to y= e^x at (3, e^3)
y - e^3 = e^3(x - 3)
y= xe^3 - 2e^3

3. Equation of normal to y=lnx at (3, ln3)
y - ln3 = -x(x - 3)
y = -x^2 + 3x + ln3

4. find nature of the stationary point x=ln2e dy/dx=2e-e^x
d2y/dx2 = -e^x
d2y/dx2 = -e^ln2e
d2y/dx2 = -2e therefore max. point

5. Estimate gradient of curve y = (x+lnx)^2 when x=3
change in y / change in x
(0.0601 + ln0.0601) / 0.01

- let me know if anything is unclear or if you want me to take a picture of my actual working. Thanks again

5. Show your steps to estimation - or give your method of doing so. Where does this info come from?

### Related university courses

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: November 7, 2016
The home of Results and Clearing

### 2,796

people online now

### 1,567,000

students helped last year
Today on TSR

### Took GCSEs this summer?

Fill in our short survey for Amazon vouchers!

### University open days

1. Bournemouth University
Wed, 22 Aug '18
2. University of Buckingham
Thu, 23 Aug '18
3. University of Glasgow
Tue, 28 Aug '18
Poll
Useful resources

### Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams