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Math Help - Further differentiation

  1. #1
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    Further differentiation

    Q1)

    Q2) Let f(x) = k^2 sin^2(2x) + bx, where x ε R and k, b are non zero
    constants
    a) Find f’(x)
    b) For x ε R, the minimum value of sin(2x)cos(2x) is
    Q3) <attached image>
    Attached Thumbnails Attached Thumbnails Further differentiation-questions.bmp  
    Last edited by 99.95; August 4th 2010 at 04:28 AM.
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  2. #2
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    1. If f(x) = \log\sqrt{1-x}, find f'(x).
    Since \log{x^n} = n\log{x}, for all n,x\in\mathbb{R}_{>0}, we have f(x) = \frac{1}{2}\log(1-x).
    Then using the chain-rule we find f'(x) = -\frac{1}{2(1-x)}.

    2. Let f(x) = k^2 sin^2(2x) + bx, where x\in\mathbb{R} and k, b are non-zero constants.
    Write it as f(x) = k^2\{\sin(2x)\}^2+bx. Then (again) by the chain-rule,
    f'(x) = (2k^2\sin{2x})(2\cos{2x})+b = 4k^2\sin{2x}\cos{2x}+b = 2k^2\sin{4x}+b.

    3. For x\in\mathbb{R}, find the minimum value of  f(x) = \sin{2x}\cos{2x}.
    Note that it's equavalent to \frac{1}{2}\sin{4x} - hence f'(x) = 2\cos{4x} and f''(x) = -8\sin{4x}. Then
    we have f'(x) = 0 \Rightarrow x = \frac{n\pi}{4}-\frac{\pi}{8}, n\in\mathbb{Z}. Recall that a minimum of f(x) is a point b such that
    f'(b) = 0 and f''(b) > 0. Take the first zero of f'(x), that is x = -\frac{\pi}{8}, and we have f''(x) = 8,
    so it is a minimum. Thus the minimum value of f(x) is \frac{1}{2}\sin\left(\frac{-4\pi}{8}\right) = -\frac{1}{2}.

    4. Find f'(x) if f(x) = e^{\frac{x}{3}}-2\log\left(\sqrt{\frac{x}{3}}\right)
    By the chain-rule (again) the derivative of [tex]e^{\frac{x}{3}}[/Math] is \frac{1}{3}e^{\frac{x}{3}} (fill the details). Since \log{x^n} = n\log{x} for any
    positive real n, we have -2\log\left(\sqrt{\frac{x}{3}}\right) = \log\left(\frac{3}{x}\right) = \log{3}-\log{x} and the derivative of that is
    (obviously) -\frac{1}{x}. Thus f'(x) = \frac{1}{3}e^{\frac{x}{3}}-\frac{1}{x}. (It's late over here so I apologise if there happen to be any
    'mistypes' or mistakes).
    Last edited by TheCoffeeMachine; August 7th 2010 at 10:31 PM.
    Thanks from 99.95
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