Advanced Higher Mathematics Help

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)

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 ^.

You should know a rule that connects $\sec^2x$ and $\tan^2x$

You should know a rule that connects $\sec^2x$ and $\tan^2x$

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

Yep, from Higher

should I take it from the line:
secx (sec^2x + tan^2x)

Is it 1 + tan^2x = sec^2x

Yep, from Higher

should I take it from the line:
secx (sec^2x + tan^2x)

Yes

It comes from

$\sin^2 x + \cos^2 x = 1$

If you divide this by $\cos^2x$ or by $\sin^2 x$ you get 2 really useful identities

Yep

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?

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)

This one is chain rule not product rule

Thanks!

How do you know when to use the chain rule and product rule?

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

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.