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    x = cox y/2 + 1

    Find dx/dy when y = 2

    I differentiated it and got -sin y/2 but apparently it's - 1/2 sin y/2 ... I don't get where the -1/2 comes from as cos differentiated is -sin and the 1 disappears...

    Probably missing something simple here but I can't see what. Thanks!
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    When differentiating a trigonometric function you need to differentiate what's within the function and multiply the result by it. In this case d/dy(y/2)= 1/2. This is where the half comes from.
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    You need to use the chain rule here. Let u=\dfrac{x}{2}.
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    (Original post by nuodai)
    You need to use the chain rule here. Let u=\dfrac{x}{2}.
    Ah, I see that, thank you for your help!

    If I could just ask something else, do you know how I would go about differentiating x ln (4x - 3) ?

    Thanks again for your help!
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    (Original post by themagicpiano)
    Ah, I see that, thank you for your help!

    If I could just ask something else, do you know how I would go about differentiating x ln (4x - 3) ?

    Thanks again for your help!
    You can use product rule where u=x and v=ln(4x-3), you know how to differentiate ln functions right?
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    (Original post by themagicpiano)
    Ah, I see that, thank you for your help!

    If I could just ask something else, do you know how I would go about differentiating x ln (4x - 3) ?

    Thanks again for your help!
    product rule!

    let u = x and v = ln(4x-3)

    then dy/dx = u(dv/dx)+v(du/dx)
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    well this is how it goes

    x=cos y/2 is the same as:

    x= cos 1/2y

    so to differentiate
    cos differentiates to -sin

    the coefficient of y (1/2) is bought to the front

    so you get dx/dy = -1/2sin y/2

    Hope this helps
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    (Original post by roar558)
    You can use product rule where u=x and v=ln(4x-3), you know how to differentiate ln functions right?
    Yeah, thank you and thank you @the greatest. I just got a bit confused about how to split them up because of the ln. It just confused me. Thanks for your help.
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    (Original post by viksta1000)
    well this is how it goes

    x=cos y/2 is the same as:

    x= cos 1/2y

    so to differentiate
    cos differentiates to -sin

    the coefficient of y (1/2) is bought to the front

    so you get dx/dy = -1/2sin y/2

    Hope this helps
    Great explanation, thank you! Makes it so much clearer!

    So the coefficient always gets moved to the front when differentiating?

    Thanks
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    (Original post by themagicpiano)
    Great explanation, thank you! Makes it so much clearer!

    So the coefficient always gets moved to the front when differentiating?

    Thanks
    only in trig functions...you can use that as a general rule yes
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    (Original post by themagicpiano)
    Great explanation, thank you! Makes it so much clearer!

    So the coefficient always gets moved to the front when differentiating?

    Thanks
    (Original post by viksta1000)
    only in trig functions...you can use that as a general rule yes
    As it happens it's fine for any function.

    Suppose instead of y=f(x) you have y=f(ax+b). Then if we set u=ax+b we have y=f(u), so \dfrac{dy}{du} = f'(u)=f'(ax+b) and \dfrac{du}{dx}=a. Putting this together with the chain rule, we get \dfrac{dy}{dx} = af'(ax+b). So, if you have a function of something in the form ax+b, you can just take the derivative as you would normally and then multiply by a.
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    (Original post by nuodai)
    As it happens it's fine for any function.

    Suppose instead of y=f(x) you have y=f(ax+b). Then if we set u=ax+b we have y=f(u), so \dfrac{dy}{du} = f'(u)=f'(ax+b) and \dfrac{du}{dx}=a. Putting this together with the chain rule, we get \dfrac{dy}{dx} = af'(ax+b). So, if you have a function of something in the form ax+b, you can just take the derivative as you would normally and then multiply by a.
    i stand corrected

    i just didn't want the OP to get confused when it comes to logs

    e.g. ln2x => 1/x or 3ln2x => 3/x etc
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    (Original post by viksta1000)
    i stand corrected

    i just didn't want the OP to get confused when it comes to logs

    e.g. ln2x => 1/2x or 3ln2x => 3/2x etc
    They still give the right answer if you do it properly though. Like \dfrac{d}{dx} (\ln 2x) = \dfrac{1}{2x} \times 2 = \dfrac{1}{x}, etc... but yes, people do often get confused by differentiating logarithms.
 
 
 
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