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# C3 Differentiation Mixed Excercise Help! watch

1. I'm stuck on part (b) question 11 in Excercise 8K. The question is as follows...

" The Curve C has the eqn y = f(x), where

f(x) = 3lnx + (1/x), x > 0

(a) Calculate the x cordinate of P. -> I've correctly identified it to be 1/3
The point point Q on C has the x co-ordinate 1.
(b) Find and equation for the normal to C at Q"

... I'm actually quite stuck on what (b) is asking for.

Clues on how I could answer this question would be greatly appreciated. Thank you.
I'm stuck on part (b) question 11 in Excercise 8K. The question is as follows...

" The Curve C has the eqn y = f(x), where

f(x) = 3lnx + (1/x), x > 0

(a) Calculate the x cordinate of P. -> I've correctly identified it to be 1/3
The point point Q on C has the x co-ordinate 1.
(b) Find and equation for the normal to C at Q"

... I'm actually quite stuck on what (b) is asking for.

Clues on how I could answer this question would be greatly appreciated. Thank you.
first differentiate the function

Spoiler:
Show
f(x) = 3lnx + x^-1

you should be able to do that
3. (Original post by Chaoslord)
first differentiate the function

Spoiler:
Show
f(x) = 3lnx + x^-1

you should be able to do that
... I'd done that for part (a), I got: 3/x - x^(-2)
4. You need to differentiate f(x) to find gradient of tangent at that point. The normals gradient is the negative reciprical of this.
Use the equation y=mx+c to do the rest.
Hope this helps.
5. oops, just relaised, i've differentiated differently to you! argh, but the i differentiated did get the the correct answer for (a)... i'm confused!
... I'd done that for part (a), I got: 3/x - x^(-2)

if you sub in an x value into this equation you will get the gradient at x

now we want the normal, the normal is perpendicular to the tangent. if two lines are perpendicular the multiple of their gradients is -1

and we can use the formula

y - y1 = m(x - x1)
oops, just relaised, i've differentiated differently to you! argh, but the i differentiated did get the the correct answer for (a)... i'm confused!
no your right i was re writing the formula to make it simpler to differentiate
8. heres a fully worked solution

Spoiler:
Show

subbing in x = 1, we can see the gradient of the tangent is 2 therefore the gradient of the normal is -1/2

subbing in the 1 into the original equation we get y = 1

using y - y1 = m(x - x1)

we get

9. (Original post by Chaoslord)

if you sub in an x value into this equation you will get the gradient at x

now we want the normal, the normal is perpendicular to the tangent. if two lines are perpendicular the multiple of their gradients is -1

and we can use the formula

y - y1 = m(x - x1)
I've subbed 1 into dy/dx, this has given me 2

would the eqn for the normal be: -(1/3)lnx + 1/x

... what would I now do? I recognise this seems like C2 work, but I'm so tired at the momment and I can't seem to think properly or correctly!
11. thanks for the spolier, I've relaised i've done the normal eqn totally wrong! i get it now! thanks x
thanks for the spolier, I've relaised i've done the normal eqn totally wrong! i get it now! thanks x
happy to health
13. ... another question, its reagardin question 15. How could simplify: 1 / (2e(ln)e + e) where e in this example is being used instead of x?
... another question, its reagardin question 15. How could simplify: 1 / (2e(ln)e + e) where e in this example is being used instead of x?

this question is messy

thats what im assuming the formula is

first you can right it like this

now the chain rule states,

we can take as one function and as the other

now you may not be able to differentiate

you have to use the product rule where 2e = g(x) and ln(e) = f(x)

hope this helps
15. I don't think i need to differentiate that as i already had, I think. If it helps, the answer should be 1/3e. How would I get to that compared to what I already had?
I don't think i need to differentiate that as i already had, I think. If it helps, the answer should be 1/3e. How would I get to that compared to what I already had?
oh sorry i miss read the question

hmm im stuck are you sure you differentiated correctly? o.O
17. (Original post by Chaoslord)
oh sorry i miss read the question

hmm im stuck are you sure you differentiated correctly? o.O
This is the whole question:

Given that x = (y^2) ln y , y > 0,

(a) find dy/dx

... I did:

let u = y^2 and v = lny
du/dy = 2y and dv/dy = 1/y
,', dx/dy = y + 2ylny

... this was correct at the back

(b) use your answer to part (a) to find in terms of e, the value of dy/dx at y = e

dy/dx = 1 dx/dy

=> 1 / (2e(lne) + e)

... and i'm stuck here! it should come to 1/3e

Nb. i'm sorry about the messiness of my eqns, i don't know how to use LaTeX
This is the whole question:

Given that x = (y^2) ln y , y > 0,

(a) find dy/dx

... I did:

let u = y^2 and v = lny
du/dy = 2y and dv/dy = 1/y
,', dx/dy = y + 2ylny

... this was correct at the back

(b) use your answer to part (a) to find in terms of e, the value of dy/dx at y = e

dy/dx = 1 dx/dy

=> 1 / (2e(lne) + e)

... and i'm stuck here! it should come to 1/3e

Nb. i'm sorry about the messiness of my eqns, i don't know how to use LaTeX
e is a constant...

i'll give you a clue lne =1 xD

jesus that was easy xD i was gonna ask you if it was e the constant but i assumed you would have informed me.... xD
19. (Original post by Chaoslord)
e is a constant...

i'll give you a clue lne =1 xD

jesus that was easy xD i was gonna ask you if it was e the constant but i assumed you would have informed me.... xD
Thank you! I'd forgotten lne = 1!!! ...That does seem very easy now!

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