MA0001/Exercise 8/tasks/2.tex

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2020-10-17 18:35:18 +02:00
\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}