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Math Help - Differentiation.

  1. #1
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    Differentiation.

    Just wanted to check if I am differentiating these functions correctly.

    1. f(x) = tan 2x
    f'(x) = sec^2x(2x)2
    f'(x) = 2sec^2x

    2. f(x) = (cosx-1)(sinx+1)
    f'(x) = -sinx(sinx+1)+cosx(cosx-1)
    f'(x) = -sin^2x-sinx+cos^2x-cosx

    3. f(x) = (cosx-sinx)^2
    f'(x) = 2(cosx-sinx)(-sinx-cosx)<br />

    4. f(x) = ln(cosx)+e^{cosx}
    f'(x) = \frac{1}{x}cosx(-sinx)+e^{cosx}(-sinx)

    If a function can be simplified even further, can you please show me.
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  2. #2
    MHF Contributor matheagle's Avatar
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    the chain rule says that the derivative of f(g(x)) is f'(g(x)) times g'(x)
    HERE are my corrections

    Quote Originally Posted by dwat View Post
    Just wanted to check if I am differentiating these functions correctly.

    1. f(x) = tan (2x)
    f'(x) = sec^2(2x)\cdot 2
    f'(x) = 2sec^2(2x)
    The (2x) cannot change


    I'd multiply this first
    2. f(x) = (\cos x-1)(\sin x+1)=\sin x \cos x+\cos x-\sin x -1
    f'(x)=-\sin^2 x +\cos^2 x-\sin x -\cos x
    which is the same as yours, but less work
    But your way was ok here

    f'(x) = -sinx(sinx+1)+cosx(cosx-1)
    f'(x) = -sin^2x-sinx+cos^2x-cosx

    3. f(x) = (cosx-sinx)^2
    f'(x) = 2(cosx-sinx)(-sinx-cosx)
    This is correct but you should multiply out the terms to see if there is a simpler form, which I doubt.

    4. f(x) = ln(\cos x)+e^{\cos x}
    The first derivative is wrong, the second is correct
    f'(x) = {-\sin x\over \cos x}+e^{\cos x}(-\sin x)

    which can be reduced to -\tan x-\sin x e^{\cos x}.


    If a function can be simplified even further, can you please show me.
    Last edited by matheagle; June 5th 2009 at 12:56 AM.
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  3. #3
    Super Member Random Variable's Avatar
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    1)  f'(x) = sec^{2}(2x)*2 = 2sec^{2}(2x)

    2) correct

    3) correct

    4)  f'(x) = \frac {1}{cos(x)} \Big(-sin(x) \Big) + e^{cos(x)}\Big(-sin(x)\Big) = -tan(x) - sin(x)e^{cos(x)}
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  4. #4
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    Quote Originally Posted by Random Variable View Post
    1)  f'(x) = sec^{2}(2x)*2 = 2sec^{2}(2x)

    2) correct

    3) correct

    4)  f'(x) = \frac {1}{cos(x)} \Big(-sin(x) \Big) + e^{cos(x)}\Big(-sin(x)\Big) = -tan(x) - sin(x)e^{cos(x)}
    OK, Thanks for the corrections. what I don't get is the last one still. I understand the second part but not the first.

    f(x) = ln(cosx) using the chain rule

    wouldn't ln be the outer function and {cosx} be the inner function?

    and if you differentiate ln = \frac {1}{x}(cosx)*-sinx

    or is this the same as f'(x) = \frac {1}{cos(x)} \Big(-sin(x) \Big)
    Last edited by dwat; June 5th 2009 at 01:30 AM.
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  5. #5
    MHF Contributor matheagle's Avatar
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    you're having trouble with the chain rule
    as I wrote above you CANNOT change the inner function g(x)

    In f(x) = \ln(\cos x) the outer function is ln and the inner is cosine

    The derivative of the log is 1 over the ENTIRE inner function giving us {1\over \cos x}

    Next multiply that by the derivative of the inner function

    the derivative of the cosine x is -\sin x.

    Next multiply these two.
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  6. #6
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    Quote Originally Posted by matheagle View Post
    you're having trouble with the chain rule
    as I wrote above you CANNOT change the inner function g(x)
    Could you please explain the procedure step by step as to how you went about using the chain rule for the ln(cosx) please?
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  7. #7
    MHF Contributor matheagle's Avatar
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    It's f'(g(x))=1/cos x times g'(x) which is -sin x.

    Better now?
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  8. #8
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    Quote Originally Posted by matheagle View Post
    you're having trouble with the chain rule
    as I wrote above you CANNOT change the inner function g(x)

    In f(x) = \ln(\cos x) the outer function is ln and the inner is cosine

    The derivative of the log is 1 over the ENTIRE inner function giving us {1\over \cos x}

    Next mutiply that by the derivative of the inner function

    the derivative of the cosine x is -\sin x.

    Next multiply these two.
    I didn't change the inner function, what I did was I only differentiated ln which is where I got \frac{1}{x} then left the inner function as {cosx} and then differentiate the inner function to -sinx. Hence where I came up with \frac{1}{x}(cosx)(-sinx)

    But I see how you did it. 1 over the inner function left untouch. I'm struggling to understand why {cosx} ends up within the differentiation of ln unless you replace the x in the ln function with the inner function.

    So basically, any ln would produce a solution with \frac{1}{x} where the inner function will always end up at the bottom replacing x e.g. ln(sinx)

    f'(x) = \frac{1}{sinx}({cosx})

    is that correct?

    BTW, thank you for your help
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  9. #9
    MHF Contributor matheagle's Avatar
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    you did change the inner function, which is cosine x not just x.
    The chain rule is just the product of two functions f' of g(x) which is one over g(x)=cos x here, then times the derivative of cos x.

    Try setting f(x)=ln x and g(x) =cos x. Obtain the derivatives and plug into f'(g(x)) and g'(x).

    and yes, the derivative of ln sin x is cot x.
    More importantly you now know the integral of tan x and cot x.
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