Collection of Maths Problems

moderately complicated derivatives

Task number: 3016

• Variant 1

  \[ \frac{2^x - 1}{3x}. \]

• Resolution

  \[ \left( \frac{2^x - 1}{3x} \right)' = \left( \frac{e^{(\ln 2) x} - 1}{3x} \right)' = \frac{(\ln 2) e^{(\ln 2) x} \cdot 3x - 3\cdot(e^{(\ln 2)x} - 1)}{9x^2} = \] \[ = \frac{\ln 2 \cdot x \cdot 2^x - 2^x + 1}{3x^2}. \] The derivative exists for all \(x \neq 0\).

• Result

  \[ \frac{\ln 2 \cdot x \cdot 2^x - 2^x + 1}{3x^2}. \] The derivative exists for all \(x \neq 0\).

• Variant 2

  \[\hbox{arctan} (\ln x).\]

• Result

  \[ \frac{1}{x(1 + \ln^2 x)}. \] The derivative exists for all \(x \in (0, \infty)\).

• Variant 3

  \[\arcsin(\sin x)) \]

• Result

  The derivative is \(1\) for \[x \in \bigcup_{k \in \Bbb Z} (-\frac{\pi}2 + 2k\pi, \frac{\pi}2 + 2k\pi)\] and is \(-1\) for \[ x \in \bigcup_{k \in \Bbb Z} (\frac{\pi}2 + 2k\pi, \frac{3\pi}2 + 2k\pi).\] At the points \(\frac{\pi}2 + k\pi\), for \(k \in \Bbb Z\) the derivative does not exist.

• Variant 4

  The function \(f\) is defined by the rules:

  • \(f(x) = x^2 \sin \left(\frac{1}{x}\right)\) for \(x \in (-\infty, 0) \cup (0, \infty)\),
  • \(f(0) = 0\).

• Resolution

  We can easily determine that \[ f'(x) = 2x \sin\left(\frac1x\right) - \cos\left(\frac1x\right) \] for \(x \in (-\infty, 0) \cup (0, \infty)\).

  Now we must only compute \(f'(0)\). Let \(g(x) = x^2\) and \(h(x) = - x^2\). We can easily see that \(g(x) \geq f(x) \geq h(x)\). Also \(g'(0) = h'(0) = 0\). From there, by the theorem about the limit of a composite function (use the definition of a derivative) we have \(f'(0) = 0\).

  Notice that the function \(f\) is continuous and is differentiable everywhere, but its derivative is not continuous.