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Math Help - Differentiating

  1. #1
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    Differentiating

    I want to integrate -ln|cos(x)|+k, I and I should end with tan(x).

    However, I'm stuck. I don't know what I need to do after -x^{-1}*|cos(x)|-ln(x)*sin(x)

    This is how far I've gotten so far:

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  2. #2
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    Hello

    Something I do not understand : the title of your post is "differentiating" and you want to integrate -ln|cos(x)|+k ?

    I think that you want to differentiate -ln|cos(x)|+k
    Differentiating ln(u(x)) gives u'(x) / u(x)
    Here it gives sin(x)/cos(x) = tan(x)
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  3. #3
    Moo
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    Hello,
    Quote Originally Posted by No Logic Sense View Post
    I want to integrate -ln|cos(x)|+k, I and I should end with tan(x).

    However, I'm stuck. I don't know what I need to do after -x^{-1}*|cos(x)|-ln(x)*sin(x)

    This is how far I've gotten so far:

    I think you have a little problem with the chain rule of differentiation.

    It is stated as below :

    [f(g(x))]'=g'(x)f'(g(x))

    To let you visualize it, if you let t=g(x), it'll be :
    [f(t)]'=g'(x)f'(t)


    So here, what do you have ?
    \ln |\cos(x)|

    so f(t)=\ln |\cos(x)|

    so f is the logarithm function.
    and t=g(x)=\cos(x)

    by using the formula, you'll have the derivative of your function :

    =\underbrace{(-\sin(x))}_{g'(x)} \cdot \overbrace{\frac{1}{t}}^{f'(t)}=-\sin(x) \frac{1}{\cos(x)}


    it's not formal, but it helps you understand how to use the formula.




    Note that you just have to multiply the final result by -1 to get the derivative you're looking for
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  4. #4
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    Visual AND formal...





    As usual, straight continuous lines differentiate with respect to x and the straight dashed line with respect to the dashed balloon expression, so that the triangular network satisfies the chain rule.

    Don't integrate - balloontegrate! Balloon Calculus: worked examples from past papers
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