Ableitungsregeln

15. Oktober 2008

Berechnen Sie die Ableitung folgender Funktionen.

f_1(x)=(2x+3)^4(4x^2-6)^7

f_2(x)=(2x^4-3)^5(3x-4x^5)^4

f_3(x)=(3x+2)^4\sqrt{1+x^2}

f_4(x)=(x+2)^2\sqrt{x+1}

f_5(x)=(x^3-1)^3\sqrt{x^2+x+1}

f_6(x)=\sqrt{\frac{x+2}{x-2}}

f_7(x)=\sqrt{\frac{1}{x^2-2}}

f_8(x)=\sqrt{1-x}

f_9(x)=\sqrt{\sin x}

f_{10}(x)=\frac{\cos x+x}{\sin x-x}

Entry Filed under: Differenzialrechnung. Schlagworte: , , , .

1 Comment Add your own

  • 1. alfredmuehlleitner  |  15. Oktober 2008 at 23:17

    Lösungen:

    f'_1(x)=24\cdot (2x+3)^3(4x^2-6)^6(6x^2+7x-2)

    f'_2(x)=4\cdot (2x^4-3)^4(3x-4x^5)^3(-80x^8+96x^4-9)

    f'_3(x)=\frac{(3x+2)^3(15x^2+2x+12)}{\sqrt{1+x^2}}

    f'_4(x)=\frac{(x+2)\, (5x+6)}{2\sqrt{x+1}}

    f'_5(x)=\frac{(x^3-1)^2(20x^4+19x^3+18x^2-2x-1)}{2\sqrt{x^2+x+1}}

    f'_6(x)=-\frac 2{(x-2)^2}\sqrt{\frac{x-2}{x+2}}

    f'_7(x)=-\frac{x\sqrt{x^2-2}}{(x^2-2)^2}

    f_8'(x)=-\frac1{2\sqrt{1-x}}

    f_9'(x)=\frac{\cos x}{2\sqrt{\sin x}}

    f_{10}'(x)=\frac{-1+\sin x+x\sin x+\cos x-x\cos x}{(\sin x-x)^2}

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