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    May sound silly question, but trying to teach myself... can someone help me with finding the derivatives of the following equations:

    e^x/2

    Ln(x/4)

    Find the gradient of the curve whose equation is y=e^3x at the point (0,1)
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    You should look into chain derivatives for compositions of functions. Otherwise, you have to rely upon a table of derivatives for the derivative of the natural logarithm.
    If you ever need examples of full solutions to derivatives, you should try out this derivative calculator with steps.
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    (Original post by Olmeister)
    May sound silly question, but trying to teach myself... can someone help me with finding the derivatives of the following equations:

    e^x/2

    Ln(x/4)

    Find the gradient of the curve whose equation is y=e^3x at the point (0,1)
    Use the chain rule.
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    As people above me said, use the chain rule... for the ln(x/4) though I would use one of the log rules first
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    (Original post by asinghj)
    As people above me said, use the chain rule... for the ln(x/4) though I would use one of the log rules first
    We want dy/dx
    y=ln(x/4)
    let u = x/4
    du/dx = 1/4
    ...
    y=ln(u)
    dy/du = 1/u
    1/u = 4/x

    therefore dy/du *du/dx = dy/dx (in this case behaves like a fraction and du cancels)
    so 4/x * 1/4 = 1/x
    dy/dx = 1/x
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    (Original post by Kawaii289)
    We want dy/dx
    y=ln(x/4)
    let u = x/4
    du/dx = 1/4
    ...
    y=ln(u)
    dy/du = 1/u
    1/u = 4/x

    therefore dy/du *du/dx = dy/dx (in this case behaves like a fraction and du cancels)
    so 4/x * 1/4 = 1/x
    dy/dx = 1/x
    I know that's works but I would do this:

     ln(\frac{x}{4}) = ln(x) - ln(4)

    Therefore

     \frac{dy}{dx} = \frac{1}{x}

    The ln(4) is a constant so it becomes 0 after differentiating
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    (Original post by asinghj)
    I know that's works but I would do this:

     ln(\frac{x}{4}) = ln(x) - ln(4)

    Therefore

     \frac{dy}{dx} = \frac{1}{x}

    The ln(4) is a constant so it becomes 0 after differentiating
    Oh I see what you mean. Yes! that is a much nicer way of doing it for this question!
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    (Original post by Kawaii289)
    Oh I see what you mean. Yes! that is a much nicer way of doing it for this question!
    I used to do it the way you did it, but my teacher showed this way and it becomes so much simpler
 
 
 
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