\( \newcommand{\br}[1]{\left( #1\right)} \newcommand{\logpar}[1]{\log\left( #1\right)} \newcommand{\cospar}[1]{\cos\left( #1\right)} \newcommand{\sinpar}[1]{\sin\left( #1\right)} \newcommand{\tanpar}[1]{\tan\left( #1\right)} \newcommand{\arcsinpar}[1]{\sin^{-1}\!\left( #1\right)} \newcommand{\arccospar}[1]{\cos^{-1}\!\left( #1\right)} \newcommand{\arctanpar}[1]{\tan^{-1}\!\left( #1\right)} \newcommand{\asin}[1]{\sin^{-1}\! #1} \newcommand{\acos}[1]{\cos^{-1}\! #1} \newcommand{\atan}[1]{\tan^{-1}\! #1} \newcommand{\asinh}[1]{\sinh^{-1}\! #1} \newcommand{\acosh}[1]{\cosh^{-1}\! #1} \newcommand{\atanh}[1]{\tanh^{-1}\! #1} \newcommand{\logten}[1]{\log_{10}\! #1} \definecolor{explaination}{RGB}{0, 166, 226} \newcommand{\ubrace}[2][u]{ { \color{explaination}{\underbrace{ {\color{black}{#2}} }_{#1}} } } \newcommand{\obrace}[2][u]{ { \color{explaination}{\overbrace{ {\color{black}{#2}} }^{#1}} } } \definecolor{highlight}{RGB}{181, 41, 118} \newcommand{\xplain}[1]{{ \textcolor{explaination} { \footnotesize{ #1 \newline}}}} \newcommand{\hilite}[1]{{ \textcolor{highlight} { { #1 }}}} \definecolor{lightergray}{gray}{.675} \newcommand{\hide}[1]{{ \textcolor{lightergray} { \footnotesize{ #1 \newline}}}} \newcommand{\mth}[1]{ { \textcolor{black} { { \small #1 } } } } \)

Übung: Produktregel

generateSpecialFunction("x") generateSpecialFunction("x") funcNotation("x")

Bestimme \displaystyle \frac{d}{dx} \left((FUNCN.fText)(FUNCD.fText)\right).

\lrsplit{(FUNCN.ddxFText)(FUNCD.fText)}{+\:(FUNCD.ddxFText)(FUNCN.fText)}

  • \lrsplit{(FUNCD.fText)(FUNCN.ddxFText)}{-\:(FUNCN.fText)(FUNCD.ddxFText)}
  • \lrsplit{(FUNCN.fText)(FUNCN.ddxFText)}{+\:(FUNCD.fText)(FUNCD.ddxFText)}
  • \lrsplit{(FUNCN.fText)(FUNCD.ddxFText)}{-\:(FUNCD.fText)(FUNCN.ddxFText)}

Wir wissen, dass \displaystyle \frac{d}{dx} \bigl( f(x) g(x) \bigr) = f'(x)g(x) + g'(x)f(x).

In diesem Fall,

\qquad f(x) = FUNCN.fText,

\qquad g(x) = FUNCD.fText.

Die beiden Funktionen ableiten:

\qquad f'(x) = FUNCN.ddxFText,

\qquad g'(x) = FUNCD.ddxFText.

Daher ist die Antwort

\qquad {(FUNCN.ddxFText)(FUNCD.fText) + (FUNCD.ddxFText)(FUNCN.fText)}.