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# Advanced Higher Mathematics Help Watch

1. Hi,

Can someone help me with the questions stated below please. I have done some working but I am finding it difficult when it comes to taking out a common factor/simplifying to match the answer in the book.

Calculate dy/dx for:

1. y = sec x tan x
2. y = cot (tan x )
3. y = cosec(sin x)

My working so far:

1. y = sec x tan x
dy/dx = secx . sec^2x + tanx . secxtanx
= sec^3x + tanx . secxtanx
= secx ( sec^2x + tan^2x)
= secx(1/cos^2x + sin^2x/cos^2x)
= secx(1 + sin^2x+/cos^2x )
= ...

Answer in the book = sec x (2 sec^2x - 1)
2. (Original post by kurt123)
Hi,

Can someone help me with the questions stated below please. I have done some working but I am finding it difficult when it comes to taking out a common factor/simplifying to match the answer in the book.

Calculate dy/dx for:

1. y = sec x tan x
2. y = cot (tan x )
3. y = cosec(sin x)

My working so far:

1. y = sec x tan x
dy/dx = secx . sec^2x + tanx . secxtanx
= sec^3x + tanx . secxtanx
= secx ( sec^2x + tan^2x)
= secx(1/cos^2x + sin^2x/cos^2x)
= secx(1 + sin^2x+/cos^2x )
= ...

Answer in the book = sec x (2 sec^2x - 1)
Are you familiar with this identity

1 + (tanx)^2 = (secx)?

Use this to replace the (tanx)^2 part I put I bold I for quote above ^.
3. (Original post by kurt123)
Hi,

Can someone help me with the questions stated below please. I have done some working but I am finding it difficult when it comes to taking out a common factor/simplifying to match the answer in the book.

Calculate dy/dx for:

1. y = sec x tan x
2. y = cot (tan x )
3. y = cosec(sin x)

My working so far:

1. y = sec x tan x
dy/dx = secx . sec^2x + tanx . secxtanx
= sec^3x + tanx . secxtanx
= secx ( sec^2x + tan^2x)
= secx(1/cos^2x + sin^2x/cos^2x)
= secx(1 + sin^2x+/cos^2x )
= ...

Answer in the book = sec x (2 sec^2x - 1)
You should know a rule that connects and
4. (Original post by krisshP)
Are you familiar with this identity

1 + (tanx)^2 = (secx)?

Use this to replace the (tanx)^2 part I put I bold I for quote above ^.
No, I only started the course 2 days ago.
However , I was told secx = 1/cosx
5. (Original post by TenOfThem)
You should know a rule that connects and
This might actually be in the A. Higher formula book.
6. (Original post by TenOfThem)
You should know a rule that connects and
Is it 1 + tan^2x = sec^2x
7. (Original post by kurt123)
No, I only started the course 2 days ago.
However , I was told secx = 1/cosx
Do you know this:

(Sinx)^2 + (Cosx)^2 =1

?

Divide both sides by (cosx)^2 and see what you get.

Another identity can be obtained if divide both sides by (sinx)^2

8. (Original post by krisshP)
Do you know this:

(Sinx)^2 + (Cosx)^2 =1

?

Divide both sides by (cosx)^2 and see what you get.

Another identity can be obtained if divide both sides by (sinx)^2

Yep, from Higher

should I take it from the line:
secx (sec^2x + tan^2x)
9. (Original post by kurt123)
Yep, from Higher

should I take it from the line:
secx (sec^2x + tan^2x)
Really I don't remember doing this at higher, then again it was a while ago now.
10. (Original post by kurt123)
Is it 1 + tan^2x = sec^2x
Yes

It comes from

If you divide this by or by you get 2 really useful identities

11. (Original post by kurt123)
Yep, from Higher

should I take it from the line:
secx (sec^2x + tan^2x)
Yep
12. (Original post by TenOfThem)
Yes

It comes from

If you divide this by or by you get 2 really useful identities

(Original post by krisshP)
Yep

Thanks very much. Finally got it.
13. For 2. y = cot(tanx) my working so far is:

dy/dx = cot . sec^2x + tanx . -cosec^2
= 1/tan . sec^2x - tanxcosec^2

Where would I go from there if that's correct?
14. (Original post by kurt123)
For 2. y = cot(tanx) my working so far is:

dy/dx = cot . sec^2x + tanx . -cosec^2
= 1/tan . sec^2x - tanxcosec^2

Where would I go from there if that's correct?
This one is chain rule not product rule
15. (Original post by kurt123)
For 2. y = cot(tanx) my working so far is:

dy/dx = cot . sec^2x + tanx . -cosec^2
= 1/tan . sec^2x - tanxcosec^2

Where would I go from there if that's correct?
Use the substitution u= tan(x)
16. (Original post by zaback21)
Use the substitution u= tan(x)
(Original post by TenOfThem)
This one is chain rule not product rule
Thanks!

How do you know when to use the chain rule and product rule?
17. (Original post by kurt123)
Thanks!

How do you know when to use the chain rule and product rule?
Because it is the function of a function not a function multiplied by a function
18. (Original post by kurt123)
Thanks!

How do you know when to use the chain rule and product rule?
When I was taught these, a good way we were taught to remember was that the product rule is as such a product (multiplication) of two functions. The chainrule on the otherhand is implemented when there is a function of a function i.e. derivative of f(g(t)). Sometimes you even have to implement both perhaps when caluclating the derivative of a product where one or more of the functions of the product has a function inside.
19. (Original post by TenOfThem)
Because it is the function of a function not a function multiplied by a function
(Original post by BAD AT MATHS)
When I was taught these, a good way we were taught to remember was that the product rule is as such a product (multiplication) of two functions. The chainrule on the otherhand is implemented when there is a function of a function i.e. derivative of f(g(t)). Sometimes you even have to implement both perhaps when caluclating the derivative of a product where one or more of the functions of the product has a function inside.
Thanks for the help.
20. I know partial fractions are meant to be easy but I am stuck on 2 questions.

Express each of the following in partial fractions.

1) 3x+1 / (x - 1) (x^2 - 1)

2) 3x + 3 / (x - 1) (x^2 + x +1)

Any help would be much appreciated.

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