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    • Thread Starter
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    how do you differentiate this??

    y=ln(x^2cosx)


    the answer is 2/x - tanx


    i keep getting stuck!!
    • Community Assistant
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    Community Assistant
    Make a substitution u = x^2 \cos{x}.

    Use the product rule to find \frac{du}{dx}

    Find \frac{dy}{du}

    Then apply the chain rule.
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    Or you could just find the differential of x^2cosx using the product rule and then note that if

     y = ln [f(x)]

     \frac{dy}{dx} = \frac{f'(x)}{f(x)}
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    split it up. Using log rules you can simplify it a lot.
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    i've split it up and found it to be

    1/x (-xsinx + 2cosx)

    but i'm not sure how to go on from there
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    Using Dad's "splitting up" method, you should have obtained:

    y=\ln(x^2 \cos{x}) = \ln x + \ln x + \ln(\cos{x})
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    Nice method.
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    v(du/dx) + u(dv/dx)
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    (Original post by insparato)
    Or you could just find the differential of x^2cosx using the product rule and then note that if

     y = ln [f(x)]

     \frac{dy}{dx} = \frac{f'(x)}{f(x)}
    Yeah, we've just been told to know that. We were taught the chain rule for half a lesson but then the said the 'or otherwise' method, written in the exam questions is quicker and easier. So just do the above and you'll get the answer straight away
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    Being able to prove that result isn't really something that's required at A level. I'm sure its a result you can obtain through differentiation by first principles with a bit of limit knowledge and algebraic manipulation.
 
 
 
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