\[f(t) = \frac{t^2 -1}{t+1} + 6t^{1/3} + \sqrt{\sin t} + 4^t\]

Jeg deriverer funksjonen ledd for ledd

Ledd 1:

\begin{align*}
  \frac{d}{dt} \frac{t^2-1}{t+1} &= \frac{d}{dt}\frac{(t+1)(t-1)}{t+1} \\
  &= \frac{d}{dt} t-1 \\
  &= 1
\end{align*}

Ledd 2:

\begin{align*}
  \frac{d}{dt} 6t^{1/3} &= \frac{1}{3} \cdot 6t^{(1/3 - 1)} \\
  &= 2t^{-2/3}\\
  &= \frac{2}{\sqrt[3]{t^2}}
\end{align*}

Ledd 3:
  
\[ \frac{d}{dt} \sqrt{\sin t} \]

$u = \sin t$

\begin{align*}
  \frac{dy}{dt} &= \frac{dy}{du} \cdot \frac{du}{dt} \\
  &= \frac{d}{du} \sqrt{u} \cdot \frac{d}{dt} \sin t \\
  &= \frac{1}{2\sqrt{u}} \cdot \cos t \\
  &= \frac{\cos t}{2\sqrt{\sin t}}
\end{align*}

Ledd 4:

\begin{align*}
  \frac{d}{dt} 4^t = 4^t ln(t) \\
\end{align*}

\[\frac{df}{dx} = 1 + \frac{2}{\sqrt[3]{t^2}} + \frac{\cos t}{2\sqrt{\sin t}} + 4^t ln(t)\]