\begin{enumerate}[label=\arabic*.]
  \item \[ f(x) = ln\left(\frac{1}{x^2}\right) \]

    La $u = \frac{1}{x^2}$

    Jeg bruker kjerneregelen:

    \begin{align*}
      \frac{dy}{dx} &= \frac{dy}{du} \cdot \frac{du}{dx} \\
      &= \frac{d}{du} ln(u) \cdot \frac{d}{dx} \frac{1}{x^2} \\
      &= \frac{1}{u} -2 \frac{1}{x^3} \\
      &= \frac{1}{\frac{1}{x^2}} -2 \frac{1}{x^3} \\
      &= x^2 -2 \frac{1}{x^3}
    \end{align*}



  \item \[g(x) = \frac{1 + \sin x}{1 + e^x + x^2}\]

  Jeg bruker kvotientregelen:
  
  $u = 1 + \sin x$

  $v = 1 + e^x + x^2$

  \[ \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{ \frac{d}{dx}(u) \cdot v - u \cdot \frac{d}{dx}(v)}{v^2} \]

  \begin{align*}
    \frac{d}{dx} u &= \frac{d}{dx} 1 + \sin x \\
    &= \cos x
  \end{align*}

  \begin{align*}
    \frac{d}{dx} v &= \frac{d}{dx} 1 + e^x + x^2 \\
    &= e^x + 2x
  \end{align*}

  \begin{align*}
    \frac{dg}{dx} &= \frac{(\cos x)(1+e^x+x^2) - (1+\sin x)(e^x + 2x)}{\left(1 + e^x + x^2\right)^2}
  \end{align*}
  
  \item \[h(x) = \sqrt{1 + \sqrt{x}}\]

    $u = 1 + \sqrt{x}$

    \begin{align*}
      \frac{dy}{dx} &= \frac{dy}{dv} \cdot \frac{du}{dx} \\
      &= \frac{d}{du} \sqrt{u} \cdot \frac{d}{dx} \left( 1 + \sqrt{x} \right) \\
      &= \frac{1}{2\sqrt{u}} \cdot \frac{1}{2\sqrt{x}} \\
      &= \frac{1}{2\sqrt{1+\sqrt{x}}} \cdot \frac{1}{2\sqrt{x}} \\
      &= \frac{1}{4\sqrt{1+\sqrt{x}} \sqrt{x}} 
    \end{align*}
\end{enumerate}