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Maths C3 - Differentiation... Help??

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\displaystyle \tan x = \frac{\sin x}{\cos x}

Then use the quotient rule.

That's probably why I don't know then, because I haven't quite covered the Quotient Rule yet hahaha :tongue:

Reply 21

RDKGames

Study Forum Helper

Original post by notnek

If you want help with these types then post some example questions.

First one is straight-forward once you know the product rule. Even without it, you can use chain rule to prove it.

Then for

y=\ln(x) \Rightarrow e^y=x

and then differentiating implictly (which you will cover in C4) gives

\frac{dy}{dx}e^y=1

\displaystyle \frac{dy}{dx}=\frac{1}{e^y}= \frac{1}{x}

As for sine going to cosine and in circles, the simplest method I can think of is using the Taylor expansion of sine, and a few other tricks, as well as using the definition of a limit, but that is beyond normal maths A-Level. Though you can of course derive the derivative of

\tan(x)

yourself once you learn the quotient rule, or even just the product rule, for that matter.

Reply 22

Philip-flop

Original post by RDKGames

First one is straight-forward once you know the product rule. Even without it, you can use chain rule to prove it.

Then for

y=\ln(x) \Rightarrow e^y=x

and then differentiating implictly (which you will cover in C4) gives

\frac{dy}{dx}e^y=1

\displaystyle \frac{dy}{dx}=\frac{1}{e^y}= \frac{1}{x}

\tan(x)

yourself once you learn the quotient rule, or even just the product rule, for that matter.

Oh wow. I may have to try prove the first one using both the chain rule and product rule later on tonight.

Thank you :smile:

Reply 23

Philip-flop

How come when I use the chain rule on the second function I get?...

v=(3x-1)^{\frac{1}{2}}

Let...

t=3x-1

\frac{dv}{dx} = \frac{1}{2} t^{-\frac{1}{2}}

\frac{dv}{dx} = \frac{1}{2} (3x-1)^{-\frac{1}{2}}

But the example says it is...

\frac{dv}{dx} = 3 \times \frac{1}{2} (3x-1)^{-\frac{1}{2}}

Where has the extra 3 come from? :frown:

(edited 7 years ago)

Reply 24

Notnek

Original post by Philip-flop

How come when I use the chain rule on the second function I get?...

v=(3x-1)^{\frac{1}{2}}

Let...

t=3x-1

\frac{dv}{dx} = \frac{1}{2} t^{-\frac{1}{2}}

\frac{dv}{dx} = \frac{1}{2} (3x-1)^{-\frac{1}{2}}

But the example says it is...

\frac{dv}{dx} = 3 \times \frac{1}{2} (3x-1)^{-\frac{1}{2}}

Where has the extra 3 come from? :frown:

As I explained above, when you use the chain rule you first make a substitution for the stuff in the brackets (say t), then you differentiate what you have now, then you multiply by the derivative of the stuff you've substituted.

More formally, it is

\displaystyle \frac{dy}{dx} = \frac{dy}{dt} \times \frac{dt}{dx}

where

\frac{dt}{dx}

is the derivative of the stuff in the brackets.

So in your example you need to multiply by the derivative of

3x-1

which is

3

Reply 25

Philip-flop

Original post by notnek

\displaystyle \frac{dy}{dx} = \frac{dy}{dt} \times \frac{dt}{dx}

where

\frac{dt}{dx}

is the derivative of the stuff in the brackets.

So in your example you need to multiply by the derivative of

3x-1

which is

3

My head feels like it might explode :frown:

So what I should be doing is....

v=(3x-1)^\frac{1}{2}

let...

t=3x-1

Differentiate this to give...

\frac{dt}{dx} = 3

... <<why do I have to differentiate here first? :/

and then how do I work out what

\frac{dv}{dx}

is from here? :frown:

edit: Wait... I think something has finally clicked. I think I understand now!

(edited 7 years ago)

bump

Original post by Philip-flop

My head feels like it might explode :frown:

So what I should be doing is....

v=(3x-1)^\frac{1}{2}

let...

t=3x-1

Differentiate this to give...

\frac{dt}{dx} = 3

... <<why do I have to differentiate here first? :/

and then how do I work out what

\frac{dv}{dx}

is from here? :frown:

edit: Wait... I think something has finally clicked. I think I understand now!

You construct

\frac{dt}{dx}

from differentiating t with respect to x. Next you find

\frac{dv}{dt}

by simply substituting your t value into the expression for v, so you get v in terms of t, which you can differentiate in terms of t thus giving dv/dt.

The reason why you find these is because

\frac{dv}{dx}

is found by doing

\frac{dv}{dt}\cdot \frac{dt}{dx}

so you need to find the two rates of change individually in order to find dv/dx.

(edited 7 years ago)

Reply 28

Philip-flop

Thanks a lot @notnek My workings for

\frac{dv}{dx}

...

(Can't believe how silly I was being considering I just learnt the chain rule last night!!)

But now I'm having trouble simplifying the f'(x) part from...

Attachment not found

...How do I put this under a common denominator...

\frac{3x^2}{2(3x-1)^{\frac{1}{2}}} + (3x-1)^{\frac{1}{2}} (2x)

C3 - Ch.8 Exa 6 -The Product Rule .png30.6KB

Reply 29

RDKGames

Study Forum Helper

Original post by Philip-flop

...How do I put this under a common denominator...

\frac{3x^2}{2(3x-1)^{\frac{1}{2}}} + (3x-1)^{\frac{1}{2}} (2x)

\displaystyle \frac{3x^2}{2(3x-1)^{\frac{1}{2}}} + (3x-1)^{\frac{1}{2}} (2x) = \frac{3x^2}{2(3x-1)^{1/2}}+\frac{2x(3x-1)^{1/2} \cdot 2(3x-1)^{1/2}}{2(3x-1)^{1/2}}

and simplify

Reply 30

Philip-flop

Original post by RDKGames

\displaystyle \frac{3x^2}{2(3x-1)^{\frac{1}{2}}} + (3x-1)^{\frac{1}{2}} (2x) = \frac{3x^2}{2(3x-1)^{1/2}}+\frac{2x(3x-1)^{1/2} \cdot 2(3x-1)^{1/2}}{2(3x-1)^{1/2}}

and simplify

Oh yeah of course!! And then I could see that as...

\frac{3x^2}{2(3x-1)^{1/2}}+\frac{2(3x-1)^1(2x)}{2(3x-1)^{1/2}}

\frac{3x^2}{2(3x-1)^{1/2}}+\frac{12x^2-4x}{2(3x-1)^{1/2}}

\frac{3x^2+12x^2-4x}{2(3x-1)^{1/2}}

and go from there :smile:

Thank you @RDKGames

Reply 31

Philip-flop

This example is about proving the Quotient Rule.
I don''t understand how the second to last line goes from that to the last line when introducing

v^2

as a common denominator :frown:

C3 Chapt.8 Exa 8 - Proving the Quotient Rule.png

Reply 32

mathcoachni

The 1/v du/dx term has to be multiplied by v to have a common denominator v^2. Becoming (v du/dx)/v^2. Then the two terms have just been swapped around as to not have a leading negative in the final solution.

Reply 33

NotNotBatman

Original post by Philip-flop

This example is about proving the Quotient Rule.
I don''t understand how the second to last line goes from that to the last line when introducing

v^2

as a common denominator :frown:

-u\frac{dv}{dx}

stays the same because the denominator is already

v^2

but

\frac{1}{v}\frac{du}{dx} \times \frac{v}{v} = \frac{v}{v^2}\frac{du}{dx}

putting it all over v^2 you have what is shown.

Reply 34

Philip-flop

Original post by NotNotBatman

-u\frac{dv}{dx}

stays the same because the denominator is already

v^2

but

\frac{1}{v}\frac{du}{dx} \times \frac{v}{v} = \frac{v}{v^2}\frac{du}{dx}

putting it all over v^2 you have what is shown.

Oh right I see. So that gives me...

\frac{dy}{dx} = \frac{v du}{v^2 dx} - \frac{u dv}{v^2 dx}

But then how do I go from this to?...

\frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}

Reply 35

NotNotBatman

Original post by Philip-flop

Oh right I see. So that gives me...

\frac{dy}{dx} = \frac{v du}{v^2 dx} - \frac{u dv}{v^2 dx}

But then how do I go from this to?...

\frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}

denominators are both v^2, so the numerators stay the same.

Reply 36

RDKGames

Study Forum Helper

Original post by Philip-flop

Oh right I see. So that gives me...

\frac{dy}{dx} = \frac{v du}{v^2 dx} - \frac{u dv}{v^2 dx}

But then how do I go from this to?...

\frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}

You divide top and bottom by the change is

x

which is

dx

Reply 37

Philip-flop

Original post by RDKGames

You divide top and bottom by the change is

x

which is

dx

OMG yes!! That makes sense now! Because dividing top and bottom by

dx

has no effect on the value of the equation (as it is like dividing by 1) it merely changes the look of the equation. Thank yooooou! :smile:

Reply 38

X_IDE_sidf

Now to add nothing to the thread:

C3 calculus in a nutshell

Product rule:

y = f(x)g(x)

\displaystyle{ \frac{dy}{dx} = \lim_{\delta x \to 0} \left(\frac{f(x + \delta x)g(x + \delta x) - f(x)g(x)}{\delta x}\right) }

\displaystyle{ \frac{dy}{dx} = \lim_{\delta x \to 0} \left(\frac{f(x + \delta x)g(x + \delta x) - f(x + \delta x)g(x) + f(x+ \delta x)g(x) - f(x)g(x)}{\delta x}\right) }

\displaystyle{ \frac{dy}{dx} = \lim_{\delta x \to 0} \left( \frac{f(x + \delta x)g(x + \delta x) - f(x + \delta x)g(x)}{\delta x} \right) + \lim_{\delta x \to 0} \left( \frac{f(x+ \delta x)g(x) - f(x)g(x)}{\delta x} \right)[br]}

Unparseable latex formula:

\displaystyle{ \frac{dy}{dx} = f(x) \lim_{\delta x \to 0} \left( \frac{g(x + \delta x) - g(x)}{\delta x} \right)} + g(x) \lim_{\delta x \to 0} \left( \frac{f(x+ \delta x) - f(x)}{\delta x} \right) }

\displaystyle{ \frac{dy}{dx} = f(x)g'(x) + f'(x)g(x) }

Chain rule:

\displaystyle{ y = f(g(x)) }

...proof is fairly arduous (just pretend they are fractions

\frac{dy}{du} \frac{du}{dx} = \frac{dy}{dx}

... this works at A-levels but bare in mind THIS IS NOT THE PROOF and they are not fractions)...

\displaystyle{ \frac{dy}{dx} = f'(g(x))g'(x) }

Lastly (from the chain rule):

\frac{d}{dx} (f(y)) = f'(y) \frac{dy}{dx}

This is all you 'need' to know in terms of the rules. The quotient rule sometimes come in handy but

Unparseable latex formula:

\frax{a}{b} = ab^{-1}

so the product rule can always be used.

The 'memorise these' derivatives are:

\displaystyle{ \frac{d}{dx} (ax^n) = nax^{n-1} }

\displaystyle{ \frac{d}{dx} (e^{x}) = e^x }

\displaystyle{ \frac{d}{dx} (\ln x) = \frac{1}{x} }

\displaystyle{ \frac{d}{dx} (\sin x) = \cos x }

\displaystyle{ \frac{d}{dx} (\cos x) = - \sin x }

(HAVEN'T INCLUDED THE ONES ON THE FORMULA SHEET)

Tips:

Try taking the natural log, in the case of exponentials for example if given the function, let's say,