Mathematics Question Thread

[RuiAce]

What would be the best way to differentiate these trig functions?

[Arisa_90]

How do know when to use chain rule?

[RuiAce]

How do know when to use chain rule?

Function of a function.

Here, the function cos then gets smacked with a square. i.e. if \(f(x) = \cos x\) and \(g(x) = x^2\), then here we have \((\cos x)^2\), which is \( f(g(x))\)

Recognising when to use the rules is a process you MUST be familiar with.

[kiwiberry]

Hi! I'd love your help with question 4)b), I'm confused with what do with the given formula

A_n is the amount they owe after the nth repayment, so if the loan is to be repaid after 288 months, then A₂₈₈=0. We can use the given formula with n=288 to find M:

The part on the left is the sum of a GP, with a=1, r=1.0075 and n=288:

[Arisa_90]

for y= cos^2t /t
Can i do this?
Y= (cost)^2 /t
Dy/dx = (2 x cos t x -sin t x 1) x t - cos^2t x 1 / t^2

[RuiAce]

for y= cos^2t /t
Can i do this?
Y= (cost)^2 /t
Dy/dx = (2 x cos t x -sin t x 1) x t - cos^2t x 1 / t^2

[Arisa_90]

I see. What do you do for a equation like this? Tanx^2. Would you use the chain rule because its to a power?  So dy/dx =
(2tanx^2 )( sec^2 x^2 )(2x)

What is the difference to tan^2x?
Do you do dy/dx = (sec^2x 2x) (2)?

Also would you do when its
 x^3tan^2  x 2x?

[RuiAce]

I see. What do you do for a equation like this? Tanx^2. Would you use the chain rule because its to a power?  So dy/dx =
(2tanx^2 )( sec^2 x^2 )(2x)

What is the difference to tan^2x?
Do you do dy/dx = (sec^2x 2x) (2)?

Also would you do when its
 x^3tan^2  x 2x?

That last question is unclear. Please insert more bracketing.

[Arisa_90]

I get it now 
Thank you RuiAce
I was wondering what would be the best process to tackle this question?

[RuiAce]

I get it now 
Thank you RuiAce
I was wondering what would be the best process to tackle this question?

Note that when x=-1, y=0. Hence the x-intercept at x=-1 is important.

Else, all we care is that:
- The curve is decreasing for all x < 0, and decreasing for all 0 < x < 1
- There is a stationary point at x = 0. (Note, if you analyse it carefully it should become clear that at x=0 we have a horizontal point of inflexion)
- The curve is increasing for all x > 1. (Note, comparing this with the fact it's decreasing for x < 1, we should be able to infer that at x=1 we have a local minimum)
- The x-intercept at x = -1 should also be a point of inflexion (Note: ORDINARY point of inflexion, not the horizontal point of inflexion like at x=0)

Provided the conditions are met, the actual shape of the graph is totally irrelevant.

[cxmplete]

how would i do question 4

[RuiAce]

how would i do question 4

[katnisschung]

what's considered a turning point?
(things like max and min stationary point obviously but are all type of inflection points counted?)

[Rathin]

what's considered a turning point?
(things like max and min stationary point obviously but are all type of inflection points counted?)

A turning point is a stationary point where the gradient of the function changes (increasing or deceasing).

NOTE:
All turning points are stationary points,
BUT Not all stationary points are turning points (e.g. horizontal point of inflexion)

[Shadowxo]

what's considered a turning point?
(things like max and min stationary point obviously but are all type of inflection points counted?)

A turning point is where the gradient changes sign - ie a local maximum or minimum, and f'(x) will =0 at a turning point.
A stationary point is a point where the gradient equals zero. This includes all turning points (maximum + minimum) and some points of inflection (the ones where the gradient equals zero). f'(x)=0 at a stationary point.
A point of inflection is where the gradient goes from increasing to decreasing, or decreasing to increasing. f''(x)=0 and for stationary points of inflection (where the gradient =0), f'(x) =0 too.

Hope this helps