Lösung 1.2:1d
Aus Online Mathematik Brückenkurs 2
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}
\Bigl(\frac{\sin x}{x}\Bigr)' &= \frac{(\sin x)'\cdot x - \sin x\cdot (x)'}{x^2}\\[5pt] &= \frac{\cos x\cdot x - \sin x\cdot 1}{x^2}\\[5pt] &= \frac{\cos x}{x} - \frac{\sin x}{x^2}\,\textrm{.} \end{align} |
It is also possible to see the expression as a product of \displaystyle \sin x and \displaystyle 1/x, and to use the product rule,
| \displaystyle \begin{align}
\Bigl(\sin x\cdot\frac{1}{x}\Bigr)' &= (\sin x)'\cdot\frac{1}{x} + \sin x\cdot\Bigl(\frac{1}{x}\Bigr)'\\[5pt] &= \cos x\cdot\frac{1}{x} + \sin x\cdot\Bigl(-\frac{1}{x^2}\Bigr)\\[5pt] &= \frac{\cos x}{x} - \frac{\sin x}{x^2}\,, \end{align} |
where we have used
| \displaystyle \Bigl(\frac{1}{x}\Bigr)' = \bigl(x^{-1}\bigr)' = (-1)x^{-1-1} = -1\cdot x^{-2} = -\frac{1}{x^2}\,\textrm{.} |
