Lösung 1.2:1d
Aus Online Mathematik Brückenkurs 2
K (Lösning 1.2:1d moved to Solution 1.2:1d: Robot: moved page) |
Version vom 14:23, 16. Sep. 2008
We have a quotient between \displaystyle \sin x and \displaystyle x, and therefore one way to differentiate the expression is to use the quotient rule:
\displaystyle \begin{align}
& \left( \frac{\sin x}{x} \right)^{\prime }=\frac{\left( \sin x \right)^{\prime }\centerdot x-\sin x\centerdot \left( x \right)^{\prime }}{x^{2}} \\
& \\
& =\frac{\cos x\centerdot x-\sin x\centerdot 1}{x^{2}}=\frac{\cos x}{x}-\frac{\sin x}{x^{2}} \\
\end{align}
It is also possible to see the expression as a product of
\displaystyle \sin x and \displaystyle \frac{1}{x}, and to use the product rule,
\displaystyle \begin{align}
& \left( \sin x\centerdot \frac{1}{x} \right)^{\prime }=\left( \sin x \right)^{\prime }\centerdot \frac{1}{x}+\sin x\centerdot \left( \frac{1}{x} \right)^{\prime } \\
& \\
& =\cos x\centerdot \frac{1}{x}+\sin x\centerdot \left( -\frac{1}{x^{2}} \right)=\frac{\cos x}{x}-\frac{\sin x}{x^{2}} \\
\end{align}
where we have used
\displaystyle \left( \frac{1}{x} \right)^{\prime }=\left( x^{-1} \right)^{\prime }=\left( -1 \right)x^{-1-1}=-1\centerdot x^{-2}=-\frac{1}{x^{2}}
