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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
  2. Offline

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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
  4. Offline

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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
  6. Offline

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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?
  9. Offline

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    (Original post by fishfingers:))
    Whats the differential of lnx?
    1/x

    which is where I got the 1/sinx from
  10. Offline

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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}
  11. Offline

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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 ??
  16. Offline

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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?
  17. Offline

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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.
  18. Offline

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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
  19. Offline

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    (Original post by fishfingers:))
    Yep thank Yoouuu x
    No problem, you are welcome.
  20. Offline

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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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