Solution 1.1:2a
From Förberedande kurs i matematik 2
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- | {{ | + | By using the rule for differentiation |
- | < | + | |
- | {{ | + | {{Displayed math||<math>\frac{d}{dx}\,x^{n}=nx^{n-1}</math>}} |
+ | |||
+ | and the fact that the expression can be differentiated term by term and that constant factors can be taken outside the differentiation, we obtain | ||
+ | |||
+ | {{Displayed math||<math>\begin{align} | ||
+ | f^{\,\prime}(x) &= \frac{d}{dx}\,\bigl(x^2-3x+1\bigr)\\[5pt] | ||
+ | &= \frac{d}{dx}\,x^2 - 3\frac{d}{dx}\,x^1 + \frac{d}{dx}\,1\\[5pt] | ||
+ | &= 2x^{2-1} - 3\cdot 1x^{1-1} + 0\\[5pt] | ||
+ | &= 2x-3\,\textrm{.} | ||
+ | \end{align}</math>}} |
Current revision
By using the rule for differentiation
\displaystyle \frac{d}{dx}\,x^{n}=nx^{n-1} |
and the fact that the expression can be differentiated term by term and that constant factors can be taken outside the differentiation, we obtain
\displaystyle \begin{align}
f^{\,\prime}(x) &= \frac{d}{dx}\,\bigl(x^2-3x+1\bigr)\\[5pt] &= \frac{d}{dx}\,x^2 - 3\frac{d}{dx}\,x^1 + \frac{d}{dx}\,1\\[5pt] &= 2x^{2-1} - 3\cdot 1x^{1-1} + 0\\[5pt] &= 2x-3\,\textrm{.} \end{align} |