Core 4 Differentiation

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    Hey all!
    I'm stuck on differentiating these questions...

    1) ln(sinx) << why can't I just use the product rule on this? Like have u=ln and v=sinx and then differentiate the two to get du/dx = 1/x and dv/dx = cosx therefore giving sinx/x + lncosx? However, in the mark scheme it says...
    1/sinx x cosx = cotx... But i don't get where they've got this from?

    2)6e^tanx... I really don't know how to approach this at all
    MARK SCHEME = 6e^tanx x sec^2x = 6e^(tanx)sec^2x

    3) square root of cos2x MARK SCHEME = 1/2(cos2x)^-1/2 x (-2sin2x)
    =-(sin2x)/square root of cos2x

    If there is any kind person out there who could explain to me how the answer has been derived please, then I would greatly appreciate it
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    because

    ln(sinx)

    is ln of sinx

    not ln time sinx
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    All of the above use the Chain Rule
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    (Original post by TenOfThem)
    All of the above use the Chain Rule
    so whats the formula for the chain rule? Because I usually just work it out without the formula although its difficult to do that with the above questions
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    (Original post by fishfingers:))
    so whats the formula for the chain rule? Because I usually just work it out without the formula although its difficult to do that with the above questions
    I am not sure why these would be harder

    You know the differential of ln and the differential of sin and you times them

    ln(sinx)

    diff the ln gives 1/sinx

    diff the sinx gives cosx

    times them

    1/sinx times cosx = cosx/sinx = cotx
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    (Original post by fishfingers:))
    Hey all!
    I'm stuck on differentiating these questions...

    1) ln(sinx) << why can't I just use the product rule on this? Like have u=ln and v=sinx and then differentiate the two to get du/dx = 1/x and dv/dx = cosx therefore giving sinx/x + lncosx? However, in the mark scheme it says...
    1/sinx x cosx = cotx... But i don't get where they've got this from?
    1) e.g. differentiate ln(cosx)
    We know differentiating  \displaystyle ln(f(x)) gives  \displaystyle \frac{f'(x)}{f(x)}

    If  \displaystyle f(x)=cosx, then  \displaystyle f'(x)=-sinx
    Hence,  \displaystyle \frac{d}{dx}(ln(cosx)) = \frac{-sinx}{cosx}=-tanx
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    (Original post by raheem94)
    1) e.g. differentiate ln(cosx)
    We know differentiating  \displaystyle ln(f(x)) gives  \displaystyle \frac{f'(x)}{f(x)}

    If  \displaystyle f(x)=cosx, then  \displaystyle f'(x)=-sinx
    Hence,  \displaystyle \frac{d}{dx}(ln(cosx)) = \frac{-sinx}{cosx}=-tanx
    I though that if you differentiate lnx you get 1/x or is that integrating it? :/
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    (Original post by TenOfThem)
    I am not sure why these would be harder

    You know the differential of ln and the differential of sin and you times them

    ln(sinx)

    diff the ln gives 1/sinx

    diff the sinx gives cosx

    times them

    1/sinx times cosx = cosx/sinx = cotx
    Whats the differential of lnx?
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    (Original post by fishfingers:))
    Whats the differential of lnx?
    1/x

    which is where I got the 1/sinx from
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    (Original post by fishfingers:))
    I though that if you differentiate lnx you get 1/x or is that integrating it? :/
    Differentiating the function  \displaystyle ln(f(x)) gives  \displaystyle \frac{f'(x)}{f(x)}

    So in the case of  \displaystyle lnx ,
     \displaystyle f(x) = x differentiating f(x) gives  \displaystyle 1 hence  \displaystyle \frac{f'(x)}{f(x)} = \frac{1}{x}
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    (Original post by fishfingers:))
    Hey all!
    I'm stuck on differentiating these questions...

    1) ln(sinx) << why can't I just use the product rule on this? Like have u=ln and v=sinx and then differentiate the two to get du/dx = 1/x and dv/dx = cosx therefore giving sinx/x + lncosx? However, in the mark scheme it says...
    1/sinx x cosx = cotx... But i don't get where they've got this from?

    2)6e^tanx... I really don't know how to approach this at all
    MARK SCHEME = 6e^tanx x sec^2x = 6e^(tanx)sec^2x

    3) square root of cos2x MARK SCHEME = 1/2(cos2x)^-1/2 x (-2sin2x)
    =-(sin2x)/square root of cos2x

    If there is any kind person out there who could explain to me how the answer has been derived please, then I would greatly appreciate it
    my answer to your 1 st question is that ln 1 is = zero, so you cant use product rule with (ln1) x (sin x) as it would just give you the wrong answer.


    havent bothered reading the rest of ure questions
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    oh and dont ask people on tsr for help as the majority of people who reply to ure thread are probably doing a level maths so they are not experts in maths and may not necessarily be giving you the correct advice, instead you should be asking your maths teacher.
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    (Original post by TenOfThem)
    I am not sure why these would be harder

    You know the differential of ln and the differential of sin and you times them

    ln(sinx)

    diff the ln gives 1/sinx

    diff the sinx gives cosx

    times them

    1/sinx times cosx = cosx/sinx = cotx
    thank you but i was just wondering how did you know that it was the chain rule? because i though that to use the chain rule there must be powers involved..

    Also, how would i work out the second question then? becuase how do you differentiate e^tanx?? Is it just sec^2xe^tanx? :confused:
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    (Original post by brownieboy)
    oh and dont ask people on tsr for help as the majority of people who reply to ure thread are probably doing a level maths so they are not experts in maths and may not necessarily be giving you the correct advice, instead you should be asking your maths teacher.
    this is an a-level maths question lov!
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    (Original post by raheem94)
    Differentiating the function  \displaystyle ln(f(x)) gives  \displaystyle \frac{f'(x)}{f(x)}

    So in the case of  \displaystyle lnx ,
     \displaystyle f(x) = x differentiating f(x) gives  \displaystyle 1 hence  \displaystyle \frac{f'(x)}{f(x)} = \frac{1}{x}
    thanks ! how do i work out question 2 ??
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    (Original post by fishfingers:))
    thanks ! how do i work out question 2 ??
    Remember differentiating  \displaystyle e^{f(x)} gives,  \displaystyle f'(x)e^{f(x)}

    In your question  \displaystyle f(x)=tanx

    Can you do it now?
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    (Original post by brownieboy)
    oh and dont ask people on tsr for help as the majority of people who reply to ure thread are probably doing a level maths so they are not experts in maths and may not necessarily be giving you the correct advice, instead you should be asking your maths teacher.
    I am doing A-Level further maths, i already have an A* in A-Level maths, TenOfThem is a teacher, so who are you referring to people who won't be giving correct advice and are not experts in maths?

    We don't have so much spare time to give people wrong advice.
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    (Original post by raheem94)
    Remember differentiating  \displaystyle e^{f(x)} gives,  \displaystyle f'(x)e^{f(x)}

    In your question  \displaystyle f(x)=tanx

    Can you do it now?

    Yep thank Yoouuu x
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    (Original post by fishfingers:))
    Yep thank Yoouuu x
    No problem, you are welcome.
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    (Original post by fishfingers:))
    Hey all!
    I'm stuck on differentiating these questions...

    1) ln(sinx) << why can't I just use the product rule on this? Like have u=ln and v=sinx and then differentiate the two to get du/dx = 1/x and dv/dx = cosx therefore giving sinx/x + lncosx? However, in the mark scheme it says...
    1/sinx x cosx = cotx... But i don't get where they've got this from?

    2)6e^tanx... I really don't know how to approach this at all
    MARK SCHEME = 6e^tanx x sec^2x = 6e^(tanx)sec^2x

    3) square root of cos2x MARK SCHEME = 1/2(cos2x)^-1/2 x (-2sin2x)
    =-(sin2x)/square root of cos2x

    If there is any kind person out there who could explain to me how the answer has been derived please, then I would greatly appreciate it
    The chain rule:
    \dfrac{dy}{dx} = \dfrac{dy}{du} \times \dfrac{du}{dx}

    1: y=\ln u ,\ u=\sin x

    2: y=6e^u ,\ u=\tan x

    3: y=u^{\frac{1}{2}} ,\ u=\cos 2x

    Whenever you have a function of a function like these examples, you can use the same technique. It can help avoid making careless mistakes if nothing else!
 
 
 
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