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Math Help - Differentiation of logarithmic function

  1. #1
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    Exclamation Differentiation of logarithmic function

    Got one question from book, i try to do but my answer is wrong...
    Differentiate the function with respect to A.
    In(cosecA + cotA)

    After differentiate cosecA+cotA i stuck.
    Answer: -cosecA
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  2. #2
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    Quote Originally Posted by sanikui View Post
    Got one question from book, i try to do but my answer is wrong...
    Differentiate the function with respect to A.
    In(cosecA + cotA)

    After differentiate cosecA+cotA i stuck.
    Answer: -cosecA
    First of all, it's \ln(\csc{A} + \cot{A}), which stands for Natural Logarithm. Write the inverse trig functions in terms of what you already know...

    So y = \ln(\frac{1}{\sin{A}} + \frac{\cos{A}}{\sin{A}}) = \ln(\frac{1 + \cos{A}}{\sin{A}}).

    You have a composition of functions, so you need to use the chain rule.

    Let u = \frac{1 + \cos{A}}{\sin{A}}, so y = \ln{u}.

    Can you go from there?

    Note, to find \frac{du}{dx} you'll have to use the Quotient rule.
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    Quote Originally Posted by Prove It View Post
    First of all, it's \ln(\csc{A} + \cot{A}), which stands for Natural Logarithm. Write the inverse trig functions in terms of what you already know...

    So y = \ln(\frac{1}{\sin{A}} + \frac{\cos{A}}{\sin{A}}) = \ln(\frac{1 + \cos{A}}{\sin{A}}).

    Mr F says: Life might be easier going one step further ..... {\color{red}= \ln (1 + \cos A) - \ln (\sin A)}.

    [snip]
    ..
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    Quote Originally Posted by sanikui View Post
    Got one question from book, i try to do but my answer is wrong...
    Differentiate the function with respect to A.
    In(cosecA + cotA)

    After differentiate cosecA+cotA i stuck.
    Answer: -cosecA
    I transform the cosec-function and the cot-function into:

    cosec(A)=\dfrac1{\sin(A)} and \cot(A)=\dfrac1{\tan(A)} Then your term becomes:

    f(A) = \ln\left(\dfrac1{\sin(A)} + \dfrac1{\tan(A)}\right) Use chain rule to differentiate:

    f'(A) = \dfrac1{\dfrac1{\sin(A)} + \dfrac1{\tan(A)}} \cdot \left(-\dfrac1{\sin^2(A)} \cdot \cos(A) - \dfrac1{\sin^2(A)}\right)

    f'(A)=\dfrac{\sin(A) \cdot \tan(A)}{\sin(A) + \tan(A)} \cdot \dfrac{-(\cos(A)+1)}{\sin^2(A)} ........ Transform \tan(A)=\dfrac{\sin(A)}{\cos(A)} which will yield:

    f'(A)=\dfrac{\sin^2(A)}{\cos(A) \cdot \left(\sin(A)+\dfrac{\sin(A)}{\cos(A)}\right)} \cdot \dfrac{-(\cos(A)+1)}{\sin^2(A)} ...... Expand the bracket and cancel out \sin^2(A)

    f'(A)=\dfrac1{\sin(A) \cdot \cos(A) + \sin(A)} \cdot \dfrac{-(\cos(A)+1)}{1}

    f'(A)=\dfrac1{\sin(A) \cdot (\cos(A) + 1)} \cdot \dfrac{-(\cos(A)+1)}{1}= -\dfrac1{\sin(A)} = - cosec(A)
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  5. #5
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    Quote Originally Posted by earboth View Post
    I transform the cosec-function and the cot-function into:

    cosec(A)=\dfrac1{\sin(A)} and \cot(A)=\dfrac1{\tan(A)} Then your term becomes:

    f(A) = \ln\left(\dfrac1{\sin(A)} + \dfrac1{\tan(A)}\right) Use chain rule to differentiate:

    f'(A) = \dfrac1{\dfrac1{\sin(A)} + \dfrac1{\tan(A)}} \cdot \left(-\dfrac1{\sin^2(A)} \cdot \cos(A) - \dfrac1{\sin^2(A)}\right)

    f'(A)=\dfrac{\sin(A) \cdot \tan(A)}{\sin(A) + \tan(A)} \cdot \dfrac{-(\cos(A)+1)}{\sin^2(A)} ........ Transform \tan(A)=\dfrac{\sin(A)}{\cos(A)} which will yield:

    f'(A)=\dfrac{\sin^2(A)}{\cos(A) \cdot \left(\sin(A)+\dfrac{\sin(A)}{\cos(A)}\right)} \cdot \dfrac{-(\cos(A)+1)}{\sin^2(A)} ...... Expand the bracket and cancel out \sin^2(A)

    f'(A)=\dfrac1{\sin(A) \cdot \cos(A) + \sin(A)} \cdot \dfrac{-(\cos(A)+1)}{1}

    f'(A)=\dfrac1{\sin(A) \cdot (\cos(A) + 1)} \cdot \dfrac{-(\cos(A)+1)}{1}= -\dfrac1{\sin(A)} = - cosec(A)
    Heavy weather, earboth .....

    Continuing on from Life might be easier going one step further:

    \frac{dy}{dA} = \frac{-\sin A}{1 + \cos A} - \frac{\cos A}{\sin A} = \frac{-\sin^2 A - \cos A - \cos^2 A}{\sin A (1 + \cos A)} = \frac{-1 - \cos A}{\sin A (1 + \cos A)} = \frac{-1}{\sin A} = - \text{cosec} A.
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  6. #6
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    Quote Originally Posted by mr fantastic View Post
    Heavy weather, earboth .....

    Continuing on from Life might be easier going one step further:

    \frac{dy}{dA} = \frac{-\sin A}{1 + \cos A} - \frac{\cos A}{\sin A} = \frac{-\sin^2 A - \cos A - \cos^2 A}{\sin A (1 + \cos A)} = \frac{-1 - \cos A}{\sin A (1 + \cos A)} = \frac{-1}{\sin A} = - \text{cosec} A.
    Correct ... but sometimes you can't see the forest because of the trees
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  7. #7
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    Yeah, i know my mistake liaw...thank all!!
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