Gemischte Aufgaben zur Ableitung von sin, cos, Wurzel und zur Kettenregel

$f\left(x\right)=\sqrt{x^3}f\backslash left(x\backslash right)=\backslash sqrt\{x^3\}$

Finde die einzelnen Funktionen.

$g(x)=\sqrt{x}g(x)=\backslash sqrt\{x\}$

$h(x)=x^3\backslash \backslash h(x)=\; x^3$

$\Rightarrow f\left(x\right)=g\left(h\left(x\right)\right)\backslash \backslash \backslash Rightarrow\; f\backslash left(x\backslash right)=g\backslash left(h\backslash left(x\backslash right)\backslash right)$

Finde die einzelnen Ableitungen.

$g^{\mathrm{\prime }}\left(x\right)=\frac{1}{2\sqrt{x}}g\text{'}\backslash left(x\backslash right)=\backslash frac1\{2\backslash sqrt\; x\}$

$h^{\mathrm{\prime }}\left(x\right)=3x^{2}h\text{'}\backslash left(x\backslash right)=3x^2$

Setze nun in die Formel der Kettenregel ein.

Am Ende könntest du noch vereinfachen.

$f(x)=\sqrt{2x^{-3}}f(x)\; =\; \backslash sqrt\{2x^\{-3\}\}$

Finde die einzelnen Funktionen.

$g(x)=\sqrt{x}g(x)\; =\; \backslash sqrt\{x\}\backslash \backslash$

$h(x)=\frac{2}{x^3}h(x)\; =\; \backslash frac\{2\}\{x^3\}\backslash \backslash$

$\Rightarrow f(x)=g(h(x))\backslash Rightarrow\; f(x)\; =\; g(h(x))$

Finde die einzelnen Ableitungen.

$g^{\mathrm{\prime }}(x)={\displaystyle \frac{1}{2\sqrt{x}}}g\text{'}(x)\; =\; \backslash dfrac\{1\}\{2\backslash sqrt\{x\}\}\backslash \backslash$

$h^{\mathrm{\prime }}(x)=-{\displaystyle \frac{6}{x^4}}h\text{'}(x)\; =\; -\backslash dfrac\{6\}\{x^4\}$

Setze nun in die Formel der Kettenregel ein.

$f\left(x\right)=e^{x^3}f\; \backslash left(\; x\; \backslash right)\; =\; e^\{x^3\}$

Finde die einzelnen Funktionen.

$g\left(x\right)=e^{x}g\; \backslash left(\; x\; \backslash right)\; =\; e^x\backslash \backslash$$h\left(x\right)=x^3\backslash \backslash h\; \backslash left(\; x\; \backslash right)\; =\; x^3\backslash \backslash$$\Rightarrow f\left(x\right)=g\left(h\left(x\right)\right)\backslash \backslash \backslash Rightarrow\; f\; \backslash left(\; x\; \backslash right)\; =\; g\; \backslash left(\; h\; \backslash left(\; x\; \backslash right)\; \backslash right)$

Bestimme die einzelnen Ableitungen.

$g^{\mathrm{\prime }}\left(x\right)=e^{x}g\text{'}\backslash left(\; x\backslash right)\; =\; e^x\backslash \backslash$$h^{\mathrm{\prime }}\left(x\right)=3x^2\backslash \backslash h\text{'}\; \backslash left(\; x\; \backslash right)\; =\; 3\; x^2$

Setze nun in die Formel der Kettenregel ein.

$f(x)=ln(x^2+4)f(x)=ln\backslash left(x^2+4\backslash right)$

Finde die einzelnen Funktionen.

Bilde die Ableitung zu den gefundenen Funktionen.

$g^{\mathrm{\prime }}\left(x\right)=\frac{1}{x}g\text{'}\backslash left(x\backslash right)\; =\; \backslash frac\{1\}\{x\}\backslash \backslash$$h^{\mathrm{\prime }}\left(x\right)=2x\backslash \backslash h\text{'}\backslash left(x\backslash right)\; =\; 2x$

Setze nun alles Benötigte in die Formel ein.

$f\left(x\right)=\cos \left(x^2\right)f\backslash left(x\backslash right)=\backslash cos\backslash left(x^2\backslash right)$

Zerlege $ff$, sodass die Kettenregel angewandt werden kann.

$g\left(x\right)=\cos \left(x\right)h\left(x\right)=x^2\text{}g\backslash left(x\backslash right)=\backslash cos\backslash left(x\backslash right)\backslash \backslash h\backslash left(x\backslash right)=x^2$

$\Rightarrow f\left(x\right)=g\left(h\left(x\right)\right)\backslash Rightarrow\; f\backslash left(x\backslash right)=g\backslash left(h\backslash left(x\backslash right)\backslash right)$

Berechne die einzelnen Ableitungen.

$g^{\mathrm{\prime }}\left(x\right)=-\sin \left(x\right)h^{\mathrm{\prime }}\left(x\right)=2xg\text{'}\backslash left(x\backslash right)=-\backslash sin\backslash left(x\backslash right)\backslash \backslash h\text{'}\backslash left(x\backslash right)=2x$

Setze nun alles in die Formel der Kettenregel ein.

$f\left(x\right)={(\sin \left(x\right))}^{2}f\backslash left(x\backslash right)=\backslash left(\backslash sin\backslash left(x\backslash right)\backslash right)^2$

Zerlege $ff$, sodass die Kettenregel angewandt werden kann.

$g\left(x\right)=x^{2}h\left(x\right)=\sin \left(x\right)g\backslash left(x\backslash right)=x^2\backslash \backslash h\backslash left(x\backslash right)=\backslash sin\backslash left(x\backslash right)\backslash \backslash$

$\Rightarrow f\left(x\right)=g\left(h\left(x\right)\right)\backslash Rightarrow\; f\backslash left(x\backslash right)=g\backslash left(h\backslash left(x\backslash right)\backslash right)$

Berechne die einzelnen Ableitungen.

$g^{\mathrm{\prime }}\left(x\right)=2\cdot x{h}^{\mathrm{\prime }}\left(x\right)=\cos \left(x\right)g\text{'}\backslash left(x\backslash right)=\{\backslash textstyle2\}\backslash cdot\{\backslash textstyle\; x\}\backslash \backslash h\text{'}\backslash left(x\backslash right)=\backslash cos\backslash left(x\backslash right)$

Setze nun alles in die Formel der Kettenregel ein.

$f^{\mathrm{\prime }}\left(x\right)=g^{\mathrm{\prime }}\left(h\left(x\right)\right)\cdot h^{\mathrm{\prime }}\left(x\right)f\text{'}\backslash left(x\backslash right)=g\text{'}\backslash left(h\backslash left(x\backslash right)\backslash right)\backslash cdot\; h\text{'}\backslash left(x\backslash right)$

Setze zunächst $g^{\mathrm{\prime }}g\text{'}$ und $h^{\mathrm{\prime }}h\text{'}$ ein.

$=2\cdot (h\left(x\right))\cdot \cos \left(x\right)=2\backslash cdot(h\{\backslash left(x\backslash right)\})\backslash cdot\backslash cos\backslash left(x\backslash right)$

Nun setze $h(x)=\sin (x)h(x)=\backslash sin(x)$ ein.

$=2\cdot \sin \left(x\right)\cdot \cos \left(x\right)=2\backslash cdot\backslash sin\backslash left(x\backslash right)\backslash cdot\backslash cos\backslash left(x\backslash right)$

$f\left(x\right)=\sin \left(\frac{1}{x}\right)f\backslash left(x\backslash right)=\backslash sin\backslash left(\backslash frac1x\backslash right)$

Zerlege $ff$, sodass die Kettenregel angewandt werden kann

$g\left(x\right)=\sin \left(x\right)h\left(x\right)=\frac{1}{x}\Rightarrow f\left(x\right)=g\left(h\left(x\right)\right)g\backslash left(x\backslash right)=\backslash sin\backslash left(x\backslash right)\backslash \backslash h\backslash left(x\backslash right)=\backslash frac1x\backslash \backslash \backslash Rightarrow\; f\backslash left(x\backslash right)=g\backslash left(h\backslash left(x\backslash right)\backslash right)$

Berechne die einzelnen Ableitungen

$g^{\mathrm{\prime }}\left(x\right)=\cos \left(x\right)h^{\mathrm{\prime }}\left(x\right)=-\frac{1}{x^2}g\text{'}\backslash left(x\backslash right)=\backslash cos\backslash left(x\backslash right)\backslash \backslash h\text{'}\backslash left(x\backslash right)=-\backslash frac1\{x^2\}$

Setze nun alles in die Formel der Kettenregel ein

$f(x)=\sin (\cos (\sin (x)))f(x)=\backslash sin(\backslash cos(\backslash sin(x)))$

Zerlege $ff$ so, dass du die Kettenregel anwenden kannst.

Man sieht, dass die Verkettung (Kompositon) der Funktionen $gg$ und $hh$ mit

$g(x)=\sin (x)h(x)=\cos (\sin (x))g(x)=\backslash sin(x)\backslash \backslash h(x)=\backslash cos(\backslash sin(x))\backslash \backslash$

gerade $ff$ ergibt.

$\Rightarrow f(x)=(g\circ h)(x)=g(h(x)))=\sin (\cos (\sin (x)))\backslash Rightarrow\; f(x)\; =\; (g\; \backslash circ\; h)(x)\; =\; g(h(x)))=\backslash sin(\backslash cos(\backslash sin(x)))$

Du siehst, dass $hh$ wiederum als eine Verkettung von zwei Funktionen geschrieben werden kann. Das wird später verwendet, um die Ableitung $h^{\mathrm{\prime }}h\text{'}$ zu bestimmen.

Nach der Kettenregel gilt dann

$f^{\mathrm{\prime }}(x)=g^{\mathrm{\prime }}(h(x))\cdot h^{\mathrm{\prime }}(x)f\text{'}(x)=g\text{'}(h(x))\backslash cdot\; h\text{'}(x)$.

$g^{\mathrm{\prime }}(h(x))g\text{'}(h(x))$ kannst du direkt bestimmen:

Bestimme Ableitung von $gg$ und setze $hh$ ein.

$g^{\mathrm{\prime }}(x)=\cos (x)g\text{'}(x)=\backslash cos(x)$

$\Rightarrow g^{\mathrm{\prime }}(h(x))=\cos (\cos (\sin (x)))\backslash Rightarrow\; g\text{'}(h(x))\; =\; \backslash cos(\backslash cos(\backslash sin(x)))$

Um $hh$ abzuleiten, benötigst du wieder die Kettenregel. Zerlege also $hh$ entsprechend in $uu$ und $vv$.

$u(x)=\cos (x)v(x)=\sin (x)u(x)=\backslash cos(x)\backslash \backslash v(x)=\backslash sin(x)\backslash \backslash$

$\Rightarrow h(x)=u\circ v=u(v(x))=\cos (\sin (x))\backslash Rightarrow\; h(x)=\; u\; \backslash circ\; v\; =\; u(v(x))=\backslash cos(\backslash sin(x))\backslash \backslash$

Berechne die Ableitungen von $uu$ und $vv$, um die Kettenregel

$h^{\mathrm{\prime }}(x)=u^{\mathrm{\prime }}(v(x))\cdot v^{\mathrm{\prime }}(x)h\text{'}(x)=u\text{'}(v(x))\backslash cdot\; v\text{'}(x)$

zu verwenden.

$u^{\mathrm{\prime }}(x)=-\sin (x)v^{\mathrm{\prime }}(x)=\cos (x)u\text{'}(x)=-\backslash sin(x)\backslash \backslash v\text{'}(x)=\backslash cos(x)$

Berechne $h^{\mathrm{\prime }}h\text{'}$.

Jetzt benutze die Kettenregel, um die Ableitung von $ff$ zu berechnen