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Thread: Differentiation.

  1. #1
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    Differentiation.

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

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

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

    $\displaystyle 3. f(x) = (cosx-sinx)^2$
    $\displaystyle f'(x) = 2(cosx-sinx)(-sinx-cosx)
    $

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

    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.

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


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

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

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

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

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


    If a function can be simplified even further, can you please show me.
    Last edited by matheagle; Jun 4th 2009 at 11:56 PM.
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  3. #3
    Super Member Random Variable's Avatar
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    1) $\displaystyle f'(x) = sec^{2}(2x)*2 = 2sec^{2}(2x) $

    2) correct

    3) correct

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

    2) correct

    3) correct

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

    $\displaystyle f(x) = ln(cosx)$ using the chain rule

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

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

    or is this the same as $\displaystyle f'(x) = \frac {1}{cos(x)} \Big(-sin(x) \Big)$
    Last edited by dwat; Jun 5th 2009 at 12:30 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 $\displaystyle f(x) = \ln(\cos x)$ the outer function is ln and the inner is cosine

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

    Next multiply that by the derivative of the inner function

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

    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 $\displaystyle ln(cosx)$ 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 $\displaystyle f(x) = \ln(\cos x)$ the outer function is ln and the inner is cosine

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

    Next mutiply that by the derivative of the inner function

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

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

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

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

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

    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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