You are Here: Home >< Maths

# Ugh. Differentiation help. watch

1. Okay. So I have these 4 questions and have no idea why my answers are wrong, howver I will only show you the first one as it take forever to write this out & I have used the same principle on all of them:
find the equation of the tangents to the curves at given points:
2x^2 +9 @ x=2
for this I did dy/dx = 2*2 x^2-1
so m = 4x
then I went back and used x=2 into the raw form
y=2x^2+9
2*(2^2) + 9
y= 8+9
y=17
therefore
17 = 4x + c
x=2
so
17 = 4*2 +c
17 = 8 + c
c=9
equation of tangent = y = 4x+9
2. You have to use y-y1 = m(x-x1) to get your equation
3. (Original post by Danny.L)
Okay. So I have these 4 questions and have no idea why my answers are wrong, howver I will only show you the first one as it take forever to write this out & I have used the same principle on all of them:
find the equation of the tangents to the curves at given points:
2x^2 +9 @ x=2
for this I did dy/dx = 2*2 x^2-1
so m = 4x

then I went back and used x=2 into the raw form
y=2x^2+9
2*(2^2) + 9
y= 8+9
y=17
therefore
17 = 4x + c
x=2
so
17 = 4*2 +c
17 = 8 + c
c=9
equation of tangent = y = 4x+9
You forgot to find the value of the differential at x = 2.
4. (Original post by Jordan97)
You have to use y-y1 = m(x-x1) to get your equation
Both methods can be used. The value of m is incorrect.
5. (Original post by morgan8002)
You forgot to find the value of the differential at x = 2.
Okay, so could someone quote me exactly where I made my mistake. It's late and I've done 25 pages of notes on differentiation, so I'm having a hard time comprehending the maths terminology as I'm so worn out.
Thanks
6. (Original post by Danny.L)
Okay, so could someone quote me exactly where I made my mistake. It's late and I've done 25 pages of notes on differentiation, so I'm having a hard time comprehending the maths terminology as I'm so worn out.
Thanks
I bold coded in my previous reply to make it clear. You didn't substitute x = 2 into the differential to get the gradient.
7. (Original post by morgan8002)
I bold coded in my previous reply to make it clear. You didn't substitute x = 2 into the differential to get the gradient.
17=4x+c
x=2
4x = 4*2
17 = 8 + c
17-8 = c
c=9
I dont understand what you mean. I thought that is putting the x value into the differentiated gradient
8. (Original post by morgan8002)
I bold coded in my previous reply to make it clear. You didn't substitute x = 2 into the differential to get the gradient.
after doing the nx^nx-1 that gives you the gradient of any point on that line. It is 4x
so I take that down and use it in the y=mx+c
17 = 4x +c
I cannot see what is wrong with this and dont understand what to do. I am seemingly getting some right and some wrong with the above method, it isnt wrong 100% of the time but neither is it right.
I am yet to be taught this, I'm learning on my own, so I could use some guidence
9. (Original post by Danny.L)
17=4x+c
x=2
4x = 4*2
17 = 8 + c
17-8 = c
c=9
I dont understand what you mean. I thought that is putting the x value into the differentiated gradient

10. (Original post by morgan8002)

yeah?
so you're saying this 8 is our new gradient? or just value of x * gradient?
I would have thought it is the result of
y=4x+c
we know x =2
and thus y=8 +c

but you're saying its
y=8x+c
so infact
17=16+c ?
c=1
y=8x +1
11. (Original post by Danny.L)
yeah?
so you're saying this 8 is our new gradient? or just value of x * gradient?
I would have thought it is the result of
y=4x+c
we know x =2
and thus y=8 +c

but you're saying its
y=8x+c
so infact
17=16+c ?
c=1
y=8x +1
The differential gives you the general equation for the gradient at any point on the line. Substituting a value of x gives the gradient at that point.

Yes, y = 8x + 1.
12. (Original post by morgan8002)
The differential gives you the general equation for the gradient at any point on the line. Substituting a value of x gives the gradient at that point.

Yes, y = 8x + 1.
Yeah, I've been doing this for 5 hours straight now so am getting groggy. It's just clicked.
The general dx/dy gives you the equation of any goven point, however the tangent isn't any given point, it is a singular value, in this case x=2 y=17
so we have to then specialise (if that's correct), the dx/dy differentiation to that specific point. I completely get it now!
13. (Original post by morgan8002)
The differential gives you the general equation for the gradient at any point on the line. Substituting a value of x gives the gradient at that point.

Yes, y = 8x + 1.
Btw, last question, I have to find the equation of a tangent to 1/(x^3), was wondering if you could help?
y=1/x^3
x= -1
I understand that we need to do
y = 1/-1^3
and thus y = -1
so our coordinates are (-1,-1)
then we differentiate to get
dx/dy
so:
1 * x^-3
-3 * x-4
then we have to times put -1 (x) into the -3/ x^4
where we get -3/1 so -3
and thus
(y) -1 = -3 +c
so c= 2?
14. (Original post by Danny.L)
Yeah, I've been doing this for 5 hours straight now so am getting groggy. It's just clicked.
The general dx/dy gives you the equation of any goven point, however the tangent isn't any given point, it is a singular value, in this case x=2 y=17
so we have to then specialise (if that's correct), the dx/dy differentiation to that specific point. I completely get it now!
Yes I'm glad you got it. The differential is general to the line. The gradient of the tangent is specific to that point on the line.
15. (Original post by Danny.L)
Btw, last question, I have to find the equation of a tangent to 1/(x^3), was wondering if you could help?
y=1/x^3
x= -1
Rearrange to get , then follow roughly the same method. Tell me if you get stuck.
16. (Original post by morgan8002)
Rearrange to get , then follow roughly the same method. Tell me if you get stuck.
Just updated my last reply to you, take a look and see if that looks right to you
I'm glad I got it too! Thanks for that btw
17. (Original post by Danny.L)
Just updated my last reply to you, take a look and see if that looks right to you
I'm glad I got it too! Thanks for that btw
Yeah it looks right.
18. (Original post by Danny.L)
Btw, last question, I have to find the equation of a tangent to 1/(x^3), was wondering if you could help?
y=1/x^3
x= -1
I understand that we need to do
y = 1/-1^3
and thus y = -1
so our coordinates are (-1,-1)
then we differentiate to get
dx/dy
so:
1 * x^-3
-3 * x-4
then we have to times put -1 (x) into the -3/ x^4
where we get -3/1 so -3
and thus
(y) -1 = -3 +c
so c= 2?
You are almost there

So y=-3x+c

Put your point into that and you will see that c=-4
19. (Original post by morgan8002)
Yeah it looks right.
Not quite
20. (Original post by TenOfThem)
You are almost there

So y=-3x+c

Put your point into that and you will see that c=-4
but now I am confused again.
to get m = -3 I had to put the x=-1 value into it
1 * x^-3 is the raw form
differentiated is:
-3 * x^-4
so differentiated is
-3/x^4
then we make it specific by replacing x with -1
this gives us
-3/-1^4
and thus -3/1 = -3
so surely our sepcific gradient for the tangent is -3, we dont need to times it by -1 again.
-3 / x^4 is the gradient of any given point on the curve.
Replace x with -1 and surely the gradient isn't
-3x it is just, -3 ? Do you see what I mean? :/

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: January 21, 2015
Today on TSR

### Three reasons you may feel demotivated right now

...and how to stay positive

### Can I get A*s if I start revising now?

Discussions on TSR

• Latest
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

## Groups associated with this forum:

View associated groups
Discussions on TSR

• Latest

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE