Solution 1.2:2a
From Förberedande kurs i matematik 2
(Difference between revisions)
(Ny sida: {{NAVCONTENT_START}} <center> Bild:1_2_2a.gif </center> {{NAVCONTENT_STOP}}) |
m |
||
(3 intermediate revisions not shown.) | |||
Line 1: | Line 1: | ||
- | {{ | + | The expression is composed of two parts: first, an outer part, |
- | < | + | |
- | {{ | + | {{Displayed math||<math>\sin \bbox[#FFEEAA;,1.5pt]{\,\phantom{x^2_2}\,}</math>}} |
+ | |||
+ | and then an inner part, <math>\bbox[#FFEEAA;,1.5pt]{\,\phantom{x^2_2}\,} = x^{2}\,</math>. | ||
+ | |||
+ | When we differentiate compound expressions, we first differentiate the outer part, | ||
+ | <math>\sin \bbox[#FFEEAA;,1.5pt]{\,\phantom{xx}\,}</math>, as if | ||
+ | <math>\bbox[#FFEEAA;,1.5pt]{\,\phantom{xx}\,}</math> were the variable that we differentiate with respect to, and then we multiply with the derivative of the inner part <math>\bigl(\bbox[#FFEEAA;,1.5pt]{\,\phantom{xx}\,}\bigr)'</math>, so that | ||
+ | |||
+ | {{Displayed math||<math>\frac{d}{dx}\,\sin \bbox[#FFEEAA;,1.5pt]{\,x^2\,} = \cos \bbox[#FFEEAA;,1.5pt]{\,x^2\,}\cdot \bigl(\bbox[#FFEEAA;,1.5pt]{\,x^2\,}\bigr)' = \cos x^2\cdot 2x\,\textrm{.}</math>}} |
Current revision
The expression is composed of two parts: first, an outer part,
\displaystyle \sin \bbox[#FFEEAA;,1.5pt]{\,\phantom{x^2_2}\,} |
and then an inner part, \displaystyle \bbox[#FFEEAA;,1.5pt]{\,\phantom{x^2_2}\,} = x^{2}\,.
When we differentiate compound expressions, we first differentiate the outer part, \displaystyle \sin \bbox[#FFEEAA;,1.5pt]{\,\phantom{xx}\,}, as if \displaystyle \bbox[#FFEEAA;,1.5pt]{\,\phantom{xx}\,} were the variable that we differentiate with respect to, and then we multiply with the derivative of the inner part \displaystyle \bigl(\bbox[#FFEEAA;,1.5pt]{\,\phantom{xx}\,}\bigr)', so that
\displaystyle \frac{d}{dx}\,\sin \bbox[#FFEEAA;,1.5pt]{\,x^2\,} = \cos \bbox[#FFEEAA;,1.5pt]{\,x^2\,}\cdot \bigl(\bbox[#FFEEAA;,1.5pt]{\,x^2\,}\bigr)' = \cos x^2\cdot 2x\,\textrm{.} |