Skillnad mellan versioner av "2.2 Lösning 7"

Versionen från 16 september 2014 kl. 16.10

\[ \Delta y = f(a + h) \, - \, fa) = (a + h)^2 - a^2 = a^2 + 2\,a\,h + h^2 - a^2 = 2\,a\,h + h^2 \]

\[ \Delta x \, = \, a + h \, - \, a \, = \, h \]

\[ {\Delta y \over \Delta x} = {2\,a\,h + h^2 \over h} = {h\,(2\,a + h) \over h} = 2\,a + h \]

@@ Rad 1: / Rad 1: @@
+::
 ::<math> \Delta y = f(a + h) \, - \, fa) = (a + h)^2 - a^2 = a^2 + 2\,a\,h + h^2 - a^2 = 2\,a\,h + h^2 </math>
 ::<math> \Delta x \, = \, a + h \, - \, a \, = \, h </math>
-::<math> {\Delta y \over \Delta x} \, = \, {f(x_1 + h) \, - \, f(x_1) \over h}  </math>
+::<math> {\Delta y \over \Delta x} = {2\,a\,h + h^2 \over h} = {h\,(2\,a + h) \over h} = 2\,a + h </math>
-Om vi tillämpar derivatans definition på <math> f(x) = x^2\, </math> kan vi skriva:
-<math> f\,'(x) = \lim_{h \to 0} {f(x+h) - f(x) \over h} = \lim_{h \to 0} {(x+h)^2 - x^2 \over h} = \lim_{h \to 0} {(x^2 + 2\,x\,h + h^2) - x^2 \over h} = \lim_{h \to 0} {2\,x\,h + h^2 \over h} = </math>
-::<math> = \lim_{h \to 0} {h\,(2\,x + h) \over h} = \lim_{h \to 0} \, (2\,x + h) =  2\,x </math>