Help with C1!

Well I'm sure every C1 module has differentiation. Normal differentiation and differentiation from first principles are two slightly different things; for one you're dealing with limits and the other you're just moving the exponents down. I'm just surprised C1 has the first principles method, not that it's difficult anyway. Just asking because with AQA first principles and normal were split between C1 and FP1.

(edited 7 years ago)

Reply 62

Original post by RDKGames

Yeah i was quite pleasantly surprised when i did the normal differentiation until my teacher pulled out from first principles :hoppy:

but its nice to get something difficult out first in c1. Then theres also something about derivative and f''(x)...

Reply 63

How does f(x+h) - f(x) / (x+h) - x becomes f(x+h) -f(x) / h ?

i dont see how the denominator cancels out with anything

Reply 64

The x's cancel out.

(x+h)-x = x-x+h

Reply 65

Original post by RDKGames

The x's cancel out.

(x+h)-x = x-x+h

ooooh i thought the -x outside was meant to be multiplied with the bracket. oh.

Reply 66

Nope. Besides, why would it be multiplied? You are finding the gradient between two points; you don't multiply anything on this stage.

Reply 67

Original post by RDKGames

Nope. Besides, why would it be multiplied? You are finding the gradient between two points; you don't multiply anything on this stage.

i wasnt going to multiply it anyways, i just thought how can u cancel it if the signs are multiplication which is in fact minus

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can u confirm for me if the first paragraph ''the derivative...'' has any other topics apart from differentiation from first principles?

Reply 68

Just gonna pretend I understood what you said.

And yes, the rates of changes as well as the second order differential.

Reply 69

Original post by RDKGames

Just gonna pretend I understood what you said.

And yes, the rates of changes as well as the second order differential.

i know it's so hard to understand and explain what goes on through my head

and im gonna pretend i understand what you said

Spoiler

Reply 70

How was I unclear for you to pretend? :facepalm:

I answered your question as in YES there are topics mentioned within the first paragraph apart from differentiation from first principles; they are rates of change and second differentials. Confirmed, as you asked.

Reply 71

Original post by RDKGames

How was I unclear for you to pretend? :facepalm:

i was directing it at the topics itself :colondollar:

oops but yes u answered my question

Reply 72

[scroll] HAIIIIII [/scroll]

so i just wanted to know why the second order derivative is d^2y/dx^2 why is the d and x to the power of 2
actually, what is d to begin with. ive done differientation but dont get why d is in the equation

Reply 73

d represents an infinitely small change. So dy means an infinitely small change in y and dx is an infinitely small change in x. As you can see, change in y over change in x gives you the gradient... or the rate of change, rather.

(edited 7 years ago)

Reply 74

To answer your first question; consider the operator

\frac{d}{dx}

which differentiates some expression to the right of it with respect to x, let's call this expression y.

First derivative:

\frac{d}{dx}(y)

Seconds derivative:

\frac{d}{dx}(\frac{d}{dx}(y))

\displaystyle =\frac{d}{dx}(\frac{dy}{dx}) = \frac{d(\frac{dy}{dx})}{dx} = \frac{ddy}{dxdx} = \frac{d^2y}{dx^2}

The reason why the

d

on the 'denominator' is not squared (I think, may get corrected on this) is because we are taking

x

twice for the second derivative but only considering one change. I'm not an expert on this notation so perhaps someone else can explain.

(edited 7 years ago)

Reply 75

Zacken

Original post by RDKGames

To answer your first questions; consider the operator

\frac{d}{dx}

which differentiates some expression to the right of it with respect to x, let's call this expression y.

First derivative:

\frac{d}{dx}(y)

Seconds derivative:

\frac{d}{dx}(\frac{d}{dx}(y))

=\frac{d}{dx}(\frac{dy}{dx}) = \frac{d(\frac{dy}{dx})}{dx} = \frac{ddy}{dxdx} = \frac{d^2y}{dx^2}

The reason why the

d

on the 'denominator' is not squared (I think, may get corrected on this) is because we are taking

x

twice for the second derivative but only considering one change. I'm not an expert on this notation so perhaps someone else can explain.

You're making the mistake of treating

d

as an actual variable. It's not.

dx

is an object, it's a thing in it's own right. It's just abbreviated using two letters instead of 1. So when you have

dx dx = dx^2

that's the whole of dx being squared because the d and the x aren't two seperate things, they are one object dx. So dx^2 = dx * dx. It's sort of like you saying that sin^2 x means s * i * n * n x just becase the sqaure is on the n, it's on the entire object sin^2, since sin is an object in its own right, just like dx.

Of course, the above is just intuition for why the notation should be like that, in reality, it's just notation and we can write it whatever way we want, there isn't any proper maths behind it.

Reply 76

Original post by Zacken

You're making the mistake of treating

d

as an actual variable. It's not.

dx

is an object, it's a thing in it's own right. It's just abbreviated using two letters instead of 1. So when you have

dx dx = dx^2

Thanks for that example, couldn't find a way to think about it differently. I blame Leibnitz for this damn notation.

Reply 77

Hello, i just want a confirmation relating to ''determining whether a stationary point is a point of inflection''

when f''(x) = 0 you need to investigate further right?
so what i do is the 'gradient table' where i find out what dy/dx is when x=-1 x=0 x=1 ect

my point is, for it to be a point of inflection, dy/dx must be positive/negative and NOT change its sign (-/+) when x=-1 and x=1... am i right?

i've read the notes in the book for ages and its worded so briefly that im not so sure if that's the rules.

Reply 78

Zacken

wot?

for a stationary point to be an inflection point, you need f'(x) = 0 and f''(x) = 0.

[and the obvious caveat for anybody who cares]

Reply 79