Solution 1.2:1d
From Förberedande kurs i matematik 2
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| - | {{ | + | We have a quotient between <math>\sin x</math> and <math>x</math>, and therefore one way to differentiate the expression is to use the quotient rule, |
| - | < | + | |
| - | {{ | + | {{Displayed math||<math>\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}</math>}} | ||
| + | |||
| + | It is also possible to see the expression as a product of <math>\sin x</math> and | ||
| + | <math>1/x</math>, and to use the product rule, | ||
| + | |||
| + | {{Displayed math||<math>\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}</math>}} | ||
| + | |||
| + | where we have used | ||
| + | |||
| + | {{Displayed math||<math>\Bigl(\frac{1}{x}\Bigr)' = \bigl(x^{-1}\bigr)' = (-1)x^{-1-1} = -1\cdot x^{-2} = -\frac{1}{x^2}\,\textrm{.}</math>}} | ||
Current revision
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{.} |
