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Maths C3 - Differentiation... Help?? watch

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    So I've embarked on the last chapter (8) of the Edexcel C3 Modular Maths Textbook. I will post here for the times that I get stuck so I don't have to keep spamming TSR with new threads
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    My first question is about the Chain rule...

    Do I have to learn this version??...
    Name:  Chain Rule.png
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    Or can I simply learn the other version which is...
     \frac{dy}{dx}= \frac{dy}{du} \times \frac{du}{dx} ??

    I'm not going to lie, I don't understand anything from the first set of formulas. As far as I'm aware, not even ExamSolutions covers it!
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    Here we go again!
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    (Original post by RDKGames)
    Here we go again!
    Just when you thought you heard the last of me :P
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    (Original post by Philip-flop)
    My first question is about the Chain rule...

    Do I have to learn this version??...
    Name:  Chain Rule.png
Views: 280
Size:  29.6 KB

    Or can I simply learn the other version which is...
     \frac{dy}{dx}= \frac{dy}{du} \times \frac{du}{dx} ??

    I'm not going to lie, I don't understand anything from the first set of formulas. As far as I'm aware, not even ExamSolutions covers it!
    You can remember the last one because the ones above it have been derived by applying the one you know to certain forms of functions without loss of generality. I'll show you how your \frac{dy}{dx}=\frac{dy}{du} \cdot \frac{du}{dx} fits into the first two definitions. Though it tells you to learn them, I never did. You just get used to them as you practice them, and you can always derive them as I will show you below.


    1. As an example, pick y=(x^2+1)^2 then we let u=x^2+1 \Rightarrow \frac{du}{dx}=2x and we also now have y=u^2 \Rightarrow \frac{dy}{du}=2u

    So, coming from the rule you know, we say \frac{dy}{dx}=\frac{dy}{du}\cdot \frac{du}{dx} = 2u \cdot 2x = 4x(x^2+1) from simplifying and reverting back to x from u

    Now let's generalise this:

    Say I have a function y=[f(x)]^n. This would link to our example above because f(x)=x^2+1 and n=2.

    Now what did we do? We set whatever is inside the bracket to some new variable, so we let u=f(x)\Rightarrow \frac{du}{dx}=f'(x) as required. And we are now left with y=u^n which we know how to differentiate with respect to u so we get \frac{dy}{du}=n\cdot u^{n-1}

    Following, again, from the equation you know we get: \frac{dy}{dx}=\frac{dy}{du} \cdot \frac{du}{dx} = nu^{n-1} \cdot f'(x)=n[f(x)]^{n-1}f'(x) as shown in the book.

    2. As an example, pick f(x)=2x^2+1 and g(x)=x^4-x^3+1 then we can let y=fg(x) which you should recognise as a composite function. It means the exact same thing as y=f[g(x)] just different notation.

    So we have y=fg(x)=2(x^4-x^3+1)^2+1 and we want to differentiate this.

    Now we let u=x^4-x^3+1 \Rightarrow \frac{du}{dx}=4x^3-3x^2

    We also have y=2u^2+1 \Rightarrow \frac{dy}{du}=4u

    So, \frac{dy}{dx}=\frac{dy}{du} \cdot \frac{du}{dx} = 4u \cdot (4x^3-3x^2) = 16x^3(x^4-x^3+1)-12x^2(x^4-x^3+1)


    Now let's generalise it.

    We have y=f[g(x)] and we let u=g(x) \Rightarrow \frac{du}{dx}=g'(x)

    And y=f(u) \Rightarrow \frac{dy}{du}=f'(u)

    So; \frac{dy}{dx}=\frac{dy}{du} \cdot \frac{du}{dx} = f'(u) \cdot g'(x) = f'[g(x)] \cdot g'(x) as shown in the book.


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    (Original post by RDKGames)
    You can remember the last one because the ones above it have been derived by applying the one you know to certain forms of functions without loss of generality. I'll show you how your \frac{dy}{dx}=\frac{dy}{du} \cdot \frac{du}{dx} fits into the first two definitions. Though it tells you to learn them, I never did. You just get used to them as you practice them, and you can always derive them as I will show you below.
    ...
    ...
    Wow, thanks for that brilliant explanation!! I may have to read it a few more times to let everything fully sink in though! But this definitely makes more sense to me now! Thank you RDKGames
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    Ok so here is my noob question of the day ....

    I'm currently watching a video on Exam Solutions about 'The Product Rule'...
    Name:  The Product Rule.png
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    ... But I've hit a brick wall. How do I differentiate the term e^3^x... as in  \frac{dv}{dx} part?
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    (Original post by Philip-flop)
    Ok so here is my noob question of the day ....

    I'm currently watching a video on Exam Solutions about 'The Product Rule'...
    Name:  The Product Rule.png
Views: 256
Size:  47.0 KB

    ... But I've hit a brick wall. How do I differentiate the term e^3^x... as in  \frac{dv}{dx} part?
    v=e^{3x} \Rightarrow \frac{dy}{dx}=3e^{3x}

    Use the chain rule with g=3x if you can't differentiate e^{3x} all in one go
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    (Original post by Philip-flop)
    My first question is about the Chain rule...

    Do I have to learn this version??...
    Name:  Chain Rule.png
Views: 280
Size:  29.6 KB

    Or can I simply learn the other version which is...
     \frac{dy}{dx}= \frac{dy}{du} \times \frac{du}{dx} ??

    I'm not going to lie, I don't understand anything from the first set of formulas. As far as I'm aware, not even ExamSolutions covers it!
    If you had something like (3x^2+3)^4, you can differentiate that using the chain rule with the longwinded approach of making a substitution.

    Or with practice you can do it the quick way in your head using the formula for the derivative of anything of the form f(x)^n. And this extends to any composition of functions fg(x).

    Now you've had an explanation, can you differentiate (3x^2+3)^4 without using a substitution?
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    (Original post by RDKGames)
    v=e^{3x} \Rightarrow \frac{dy}{dx}=3e^{3x}

    Use the chain rule with g=3x if you can't differentiate e^{3x} all in one go
    I still don't think I understand

    I know that to differentiate you have to do  \frac{dy}{dx} = na^{n-1} but I don't know how to use it on the  e^{3x}

    taking your advice... I'm not sure I know how to apply the chain rule to it either :/
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    (Original post by Philip-flop)
    I still don't think I understand

    I know that to differentiate you have to do  \frac{dy}{dx} = na^{n-1} but I don't know how to use it on the  e^{3x}

    taking your advice... I'm not sure I know how to apply the chain rule to it either :/
    y=e^{3x}

    Let u=3x \Rightarrow \frac{du}{dx}=3 also y=e^u \Rightarrow \frac{dy}{du}=e^u

    So \frac{dy}{dx}=...
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    (Original post by notnek)
    If you had something like (3x^2+3)^4, you can differentiate that using the chain rule with the longwinded approach of making a substitution.

    Or with practice you can do it the quick way in your head using the formula for the derivative of anything of the form f(x)^n. And this extends to any composition of functions fg(x).

    Now you've had an explanation, can you differentiate (3x^2+3)^4 without using a substitution?
    Ok so...

    If...  y=(3x^2+3)^4

    then...  f(x) = 3x^2+3

    To differentiate using the chain rule...  [f(x)]^n = n[f(x)]^{n-1}f'(x)
    this gives ....
    4[f(x)]^3f'(x)

    4(3x^2+3)^3(6x)

    Am I right?
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    (Original post by Philip-flop)
    Ok so...

    If...  y=(3x^2+3)^4

    then...  f(x) = 3x^2+3

    To differentiate using the chain rule...  [f(x)]^n = n[f(x)]^{n-1}f'(x)
    this gives ....
    4[f(x)]^3f'(x)

    4(3x^2+3)^3(6x)

    Am I right?
    Correct. Basically the chain rule is used when you have a function of a function. In reality that just means that there are brackets somewhere. So all of these can be differentiated using the chain rule:

    (3x^2+2)^3

    e^{(3x)}

    \sin(x^2)


    Chain rule in words : Make a substitution for the stuff in the brackets then differentitate what you have now and then multiply by the derivative of the stuff in the brackets.

    So (3x^2+2)^3 :

    Start by differentiating t^3 which is 3t^2 then multiply by the derivative of 3x^2+2.

    So you're left with 3(3x^2+2)^2 \times 6x


    Next use the same method for e^{(3x)}:

    First differentiate e^t which is e^t (this is one of the standard derivatives that you'll need to learn).

    Then multiply this by the derivative of 3x.

    So you get e^{3x} \times 3.


    There are standard results that are useful to learn e.g. the derivative of e^{ax} is ae^{ax}. But you can derive all these quickly using the chain rule.
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    (Original post by RDKGames)
    y=e^{3x}

    Let u=3x \Rightarrow \frac{du}{dx}=3 also y=e^u \Rightarrow \frac{dy}{du}=e^u

    So \frac{dy}{dx}=...
    Oh right I see!

    I now understand that...
     y = ae^x

    when differentiating gives...
     \frac{dy}{dx} = ae^x ... where a is a constant.

    I think you've cleared this up now. Thanks for the help
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    (Original post by notnek)
    Correct. Basically the chain rule is used when you have a function of a function. In reality that just means that there are brackets somewhere. So all of these can be differentiated using the chain rule:

    (3x^2+2)^3

    e^{(3x)}

    \sin(x^2)


    Chain rule in words : Make a substitution for the stuff in the brackets then differentitate what you have now and then multiply by the derivative of the stuff in the brackets.

    So (3x^2+2)^3 :

    Start by differentiating t^3 which is 3t^2 then multiply by the derivative of 3x^2+2.

    So you're left with 3(3x^2+2)^2 \times 6x


    Next use the same method for e^{(3x)}:

    First differentiate e^t which is e^t (this is one of the standard derivatives that you'll need to learn).

    Then multiply this by the derivative of 3x.

    So you get e^{3x} \times 3.


    There are standard results that are useful to learn e.g. the derivative of e^{ax} is ae^{ax}. But you can derive all these quickly using the chain rule.
    Thank you notnek!! I'm still having trouble with the derivatives of exponential, natural logs, and trigonometric types. I know how to get there but I just don't fully understand how they are derived (if that makes sense?). For now I will just have to remember them and keep practising until everything clicks..

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    (Original post by Philip-flop)
    Thank you notnek!! I'm still having trouble with the derivatives of exponential, natural logs, and trigonometric types. I know how to get there but I just don't fully understand how they are derived (if that makes sense?). For now I will just have to remember them and keep practising until everything clicks..

    Name:  Differentiation Exponential, Log, and Trig types.png
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    If you want help with these types then post some example questions.
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    (Original post by notnek)
    If you want help with these types then post some example questions.
    I just don't understand how the examples in the red boxes above differentiate to what they are, which is making it more difficult to remember them. But I will attempt some more questions on these early tomo morning and see whether I have any more questions
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    (Original post by Philip-flop)
    I just don't understand how the examples in the red boxes above differentiate to what they are, which is making it more difficult to remember them. But I will attempt some more questions on these early tomo morning and see whether I have any more questions
    Oh I see. All of these are standard results that you need to learn for C3 but don't need to know where they come from.

    There is no rule e.g. the chain rule to derive these results. They can be derived from first principles of calculus. You should google them if you want more explanation but you only need to quote them for C3 without proof.

    You do however need to use the chain rule with these standard results. E.g. you need to be able to find the derivative of something like \sin \left( x^2\right) using the fact that the derivative of \sin x is \cos x.


    EDIT : The derivative of \tan x can be derived from the derivatives of \sin x and \cos x. You may be asked to show this in a C3 exam.
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    (Original post by notnek)
    Oh I see. All of these are standard results that you need to learn for C3 but don't need to know where they come from.

    There is no rule e.g. the chain rule to derive these results. They can be derived from first principles of calculus. You should google them if you want more explanation but you only need to quote them for C3 without proof.

    You do however need to use the chain rule with these standard results. E.g. you need to be able to find the derivative of something like \sin \left( x^2\right) using the fact that the derivative of \sin x is \cos x.


    EDIT : The derivative of \tan x can be derived from the derivatives of \sin x and \cos x. You may be asked to show this in a C3 exam.
    That's good then!!

    Thanks again

    Oh really? I may need to make note of how tan x is derived from the derivatives of sin x and cos x then. Once I figure out how how it is derived
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    (Original post by Philip-flop)
    That's good then!!

    Thanks again

    Oh really? I may need to make note of how tan x is derived from the derivatives of sin x and cos x then. Once I figure out how how it is derived
    \displaystyle \tan x = \frac{\sin x}{\cos x}

    Then use the quotient rule.
 
 
 
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