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generateSpecialFunction("x") generateSpecialFunction("x") FUNCN.fText FUNCN.ddxFText FUNCD.fText FUNCD.ddxFText funcNotation("x") function(a, b, c, d, e, min) { var term1 = "\\left(" + a + "\\right)" + (a === b ? "^2" : "\\left(" + b + "\\right)"); var term2 = "\\left(" + c + "\\right)" + (c === d ? "^2" : "\\left(" + d + "\\right)"); return "\\dfrac{" + term1 + min + term2 + "}" + "{\\left(" + e + "\\right)^2}"; }

Bestimme \displaystyle \frac{\mathrm{d}}{\mathrm{d}x}\biggl( \frac{FUNCN.fText}{FUNCD.fText} \biggr).

ANSWER( N_DF, D_F, D_DF, N_F, D_F, "-" )

  • ANSWER( N_DF, D_DF, D_F, N_F, D_F, "-" )
  • ANSWER( N_DF, D_F, D_DF, N_F, N_F, "-" )
  • ANSWER( N_DF, D_DF, D_F, N_F, N_F, "-" )
  • ANSWER( N_DF, D_F, D_DF, N_F, D_F, "+" )
  • ANSWER( N_DF, D_DF, D_F, N_F, D_F, "+" )
  • ANSWER( N_DF, D_F, D_DF, N_F, N_F, "+" )
  • ANSWER( N_DF, D_DF, D_F, N_F, N_F, "+" )

Von der Kettenregel und der Produktregel wissen wir, dass \displaystyle \frac{\mathrm{d}}{\mathrm{d}x\strut}\frac{f(x)}{g(x)} = \frac{f'(x)g(x) - g'(x)f(x)}{g(x){}^2}.

In diesem Fall ist

\qquad f(x) = FUNCN.fText,

\qquad g(x) = FUNCD.fText.

Beide Funktionen ableiten:

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

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

Daher ist die Antwort

\qquad \dfrac{{(FUNCN.ddxFText)(FUNCD.fText) - (FUNCD.ddxFText)(FUNCN.fText)}}{(FUNCD.fText)^2}.