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Maths C3 - Differentiation... Help?? watch

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    (Original post by notnek)
    \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
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    (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 so \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.
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    (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 so \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.
    Oh wow. I may have to try prove the first one using both the chain rule and product rule later on tonight.

    Thank you
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    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?

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    (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?
    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.
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    (Original post by notnek)
    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.
    My head feels like it might explode

    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?


    edit: Wait... I think something has finally clicked. I think I understand now!
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    (Original post by Philip-flop)
    My head feels like it might explode

    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?


    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.
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    Thanks a lot notnek My workings for  \frac{dv}{dx} ...

    Name:  Photo 11-11-2016, 13 23 48.jpg
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    (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 594200594202


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

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

    Thank you RDKGames
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    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

    Name:  C3 Chapt.8 Exa 8 - Proving the Quotient Rule.png
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    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.
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    (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

    Name:  C3 Chapt.8 Exa 8 - Proving the Quotient Rule.png
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     -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.
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    (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}
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    (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.
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    (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
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    (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!
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    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)

}

    \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 \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, y = x^{x}. This may look to do, but is actually relatively trivial.

    \displaystyle{ \ln y = x \ln x }

    \displaystyle{ \frac{1}{y} \frac{dy}{dx} = 1 + \ln x }

    \displaystyle{ \frac{dy}{dx} = x^x(1 + \ln x)}

    (now try \frac{d}{dx} \left(x^{x^x}\right), you should be able too)


    That is about it, (these are somewhat a cut and paste from my notes so sorry) hope it helped make things clearer.
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    (Original post by X_IDE_sidf)
    Now to add nothing to the thread:

    C3 calculus in a nutshell...
    ....
    ....
    Thanks for that. Your notes might actually come in handy for me
    I just wish you would give me a heads up before spamming
 
 
 
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