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    erm,

    d/dx ln | sinx | = cot x


    :o:
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    Chain rule?
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    yeah, but will u go through it? I know u guys aren't meant to tell us the answers but if i can do this 1..then i will be able to do the rest, hopefully. cheers
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    aah got it..i was getting confused with integrastion.
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    (Original post by FoOtYdUdE)
    erm,

    d/dx ln | sinx | = cot x


    :o:
    Let's break this up shall we?
    u = sin x
    v = ln |u|

    Let's find \frac{dv}{du} and \frac{du}{dx}:
    \frac{dv}{du} = \frac{1}{u} = \frac{1}{sin x}
    and
    \frac{du}{dx} = cos x

    And now to find \frac{dy}{dx} you multiply them together.
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    OK If you differentiate ln (x ) wrt x you get 1/x
    If you differentiate sin (x) wrt x you get cos x

    You have to differentiate a function-of-a-function. First take x, then sin() it. Then ln() the result.

    You start on the outside function - ln. And differentiate that. So 1/sin(x).

    Then you go inside one layer. You find sin(x). So you differentiate that. cos(x). So we've written 1/sin(x) times cos(x).

    Now we've only got x inside the sin, so we can stop.

    Then cos(x)/sin(x) = cot(x) like 1/tan.

    So to differentiate (yourgranny(x))^2 is just 2.yourgranny(x).d/dx(yourgranny(x))
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    The other way to do it is:
    d/dx ln(f(x)) = f'(x)/f(x)
    if that notation makes any sense.
 
 
 
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