inara_30
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please could someone help me with this question
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inara_30
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(Original post by inara_30)
please could someone help me with this question
i'm confused with where to differentiate because i though tany differentiated is sec2y ?
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mqb2766
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(Original post by inara_30)
i'm confused with where to differentiate because i though tany differentiated is sec2y ?
Only if you differentiate with respect to y. Youre differentiating (implicitly) with respect to x.
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(Original post by mqb2766)
Only if you differentiate with respect to y. Youre differentiating (implicitly) with respect to x.
ah ok so when differentiating implicitly i got:
x^2sec^2ydy/dx + 2xtany = 0
so dy/dx = -2xtany / x^2sec^2y
i know that 1 + tan^2y = sec^2y so i substituted that leaving with:
dy/dx = -2xtany / x^2 + x^2tan^2y
the question already says x^2tan^2y = 9 so i substituted that leaving me with:
dy/dx = -2xtany / x^2 + 9
and now i'm stuck ....
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mqb2766
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(Original post by inara_30)
ah ok so when differentiating implicitly i got:
x^2sec^2ydy/dx + 2xtany = 0
so dy/dx = -2xtany / x^2sec^2y
i know that 1 + tan^2y = sec^2y so i substituted that leaving with:
dy/dx = -2xtany / x^2 + x^2tan^2y
the question already says x^2tan^2y = 9 so i substituted that leaving me with:
dy/dx = -2xtany / x^2 + 9
and now i'm stuck ....
Bold is not quite what the question says, but get that corrected and you should be ok. You should substitute for tan(y) on the numerator and denominator.
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(Original post by mqb2766)
Bold is not quite what the question says, but get that corrected and you should be ok. You should substitute for tan(y) on the numerator and denominator.
ah silly me
ok so if i rewrite that
dy/dx = -2xtany / x^2 + tany(x^2tany)
now substitute x^2tany = 9 into that leaving with:
dy/dx = -2xtany / x^2 + 9tany
i then rearranged the question so that tany = 9/x^2 and substituted that so:
dy/dx = (-2x x 9/x^2) / (x^2 + 9(9/x^2))
which i simplify into to :
dy/dx = -18x / x^4 + 81
is that right now ?
thanks for your help
but for part b i'm really confused how do you know which number to select either side of the value given in this question to prove it's an inflection point ...
i know that an inflection point is when there's a change in sign but no matter what numbers i use it always comes out as a positive...
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mqb2766
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(Original post by inara_30)
ah silly me
ok so if i rewrite that
dy/dx = -2xtany / x^2 + tany(x^2tany)
now substitute x^2tany = 9 into that leaving with:
dy/dx = -2xtany / x^2 + 9tany
i then rearranged the question so that tany = 9/x^2 and substituted that so:
dy/dx = (-2x x 9/x^2) / (x^2 + 9(9/x^2))
which i simplify into to :
dy/dx = -18x / x^4 + 81
is that right now ?
thanks for your help
but for part b i'm really confused how do you know which number to select either side of the value given in this question to prove it's an inflection point ...
i know that an inflection point is when there's a change in sign but no matter what numbers i use it always comes out as a positive...
For a) as noted in another thread, if you did
tan(y) = 9/x^2
It would simplify the derivation a bit, but what you have done is fine. Sometimes its worth spending a minute or two seeing if a transformation may help.

For b) inflection is when the second derivative changes sign. So not exactly sure what you've done?
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inara_30
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(Original post by mqb2766)
For a) as noted in another thread, if you did
tan(y) = 9/x^2
It would simplify the derivation a bit, but what you have done is fine. Sometimes its worth spending a minute or two seeing if a transformation may help.

For b) inflection is when the second derivative changes sign. So not exactly sure what you've done?
yes i realised that after i posted but my d^2y/dx^2 has come out as:
54(x^4 - 27) / (x^4 + 81)^2
as there is x^4 - 27 on the numerator how do i prove that it's an inflection point do i just show:
x^4 - 27 = 0
x^4 = 27
therefore x = 4root27 and so it's a point of inflection ??
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mqb2766
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(Original post by inara_30)
yes i realised that after i posted but my d^2y/dx^2 has come out as:
54(x^4 - 27) / (x^4 + 81)^2
as there is x^4 - 27 on the numerator how do i prove that it's an inflection point do i just show:
x^4 - 27 = 0
x^4 = 27
therefore x = 4root27 and so it's a point of inflection ??
Pretty much. If the numerator is zero (and the denominator isn't) then the expression is zero. Strictly speaking you should show the sign changes at that point, otherwise you could have something like
y = x^4
which has a zero second derivative at 0 but no sign change so no point of inflection.
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(Original post by mqb2766)
Pretty much. If the numerator is zero (and the denominator isn't) then the expression is zero. Strictly speaking you should show the sign changes at that point, otherwise you could have something like
y = x^4
which has a zero second derivative at 0 but no sign change so no point of inflection.
ok how will i be able to show the sign changes from this second derivative - i'm confused about that
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mqb2766
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(Original post by inara_30)
ok how will i be able to show the sign changes from this second derivative - i'm confused about that
Thought it would be fairly easy?
The denominator is positive.
What is the sign of the numerator when x is greater than or less than this value. So ...
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inara_30
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(Original post by mqb2766)
Thought it would be fairly easy?
The denominator is positive.
What is the sign of the numerator when x is greater than or less than this value. So ...
the sign of the numerator will be positive so ah ok there will be a point of inflection as it would be negative as well
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mqb2766
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(Original post by inara_30)
the sign of the numerator will be positive so ah ok there will be a point of inflection as it would be negative as well
I think what you mean is the (numerator) second derivative is negative when x^4<27 and positive when x^4>27?
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inara_30
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(Original post by mqb2766)
I think what you mean is the (numerator) second derivative is negative when x^4<27 and positive when x^4>27?
ah omg yes that makes so much more sense now thank you so much
i'm assuming you're in year 13 as well so good luck with exams if you have any !
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inara_30
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mqb2766 hello sorry to bother you but you're the only one that seems to be active in this maths forum that's why i'm, asking you questions
for this question please check if i have got the write answer :
a). A = 3, B = 4, C=-2
b). attached working for part b
thanks
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mqb2766
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(Original post by inara_30)
mqb2766 hello sorry to bother you but you're the only one that seems to be active in this maths forum that's why i'm, asking you questions
for this question please check if i have got the write answer :
a). A = 3, B = 4, C=-2
b). attached working for part b
thanks
Both look good. Id have written the C term as
+ 2/(2x-1)
to keep things a bit simpler, but your working is fine.
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inara_30
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(Original post by mqb2766)
Both look good. Id have written the C term as
+ 2/(2x-1)
to keep things a bit simpler, but your working is fine.
yay thank you !!
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mqb2766
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(Original post by inara_30)
yay thank you !!
Glad your happy and

https://www.wolframalpha.com/input/?i=partial+fractions+%28-6x%5E2--11x--1%29%2F%28%28x-3%29%281-2x%29%29

Means I can be lazy ...
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inara_30
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(Original post by mqb2766)
Glad your happy and

https://www.wolframalpha.com/input/?...9%281-2x%29%29

Means I can be lazy ...
oh wow thanks for this website
i didn't know there will be one to solve partial fractions
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mqb2766
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(Original post by inara_30)
oh wow thanks for this website
i didn't know there will be one to solve partial fractions
Wolfram alpha can do most things, however copying the + into an url is a mystery.
Give me 5 to sort it out.
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