Ableitung: Hoffe ich koennt mir helfen :*

Ableitungen: 

Hallo Gast!

A) f(x)=e^x*cos(x)

$\color{blue}Produktregel\ (uv)'=u'v+uv'\\ f(x)=e^x\cdot cos\ x\\ f'(x)=e^x\cdot cos\ x+e^x\cdot (-sin\ x)\\ \color{blue}f'(x)=e^x(cos\ x-sin\ x)$

B) f(x)= ln(x+wurzel(x^2+1))

$f(x)=g(h(x))\\ Kettenregel\ f'(x)=g'(h(x))\cdot h'(x)$

$f(x)=ln(x+\sqrt{x^2+1}\ )=ln(x+(x^2+1)^{\frac{1}{2}}\\ f'(x)=\frac{1}{x+\sqrt{x^2+1}}\cdot (1+\frac{1}{2}(x^2+1)^{-\frac{1}{2}}\cdot 2x)\\ f'(x)=\frac{1+ \frac{x}{\sqrt{x^2+1}}}{x+\sqrt{x^2+1}}=\frac{\frac{\sqrt{x^2+1}+x}{\sqrt{x^2+1}}}{x+\sqrt{x^2+1}} =\frac{\sqrt{x^2+1}+x}{\sqrt{x^2+1}(x+\sqrt{x^2+1})}\\ \color{blue}f'(x)=\frac{1}{\sqrt{x^2+1}}$

C) f(x)= 3th wurzel(1+((sin(x)^2)/x)

$f(x)=\sqrt[3]{1+\frac{sin^2\ x}{x}}=(1+\frac{sin^2\ x}{x})^{\frac{1}{3}}\\ \color{blue}Quotientenregel \ f(x)=\frac{u(x)}{v(x)} \ f'(x)=\frac{u'v-uv'}{v^2}\\ $

$\large f'(x)=\frac{1}{3}(1+\frac{sin^2\ x}{x})^{-\frac{2}{3}}\cdot \frac{2\ sinx\cdot cosx\cdot x-sin^2x}{x^2}$

$\large f'(x)=\frac{2\cdot x\cdot sinx\cdot cosx-sin^2x}{3\cdot x^2\cdot \sqrt[3]{(1+\frac{sin^2\ x}{x})^2}}$

D) f(x)= ln((wurzel(ax+b)-wurzel(b)) / (wurzel(ax+b)+wurzel(b)))

                          u

$f(x)=ln(\frac{\sqrt{ax+b}-\sqrt{b}}{\sqrt{ax+b}+\sqrt{b}})$

                          v

$\color{blue}Quotientenregel \ f(x)=\frac{u(x)}{v(x)} \ f'(x)=\frac{u'v-uv'}{v^2}$

$f'(x)=\frac{\sqrt{ax+b}+\sqrt{b}}{\sqrt{ax+b}-\sqrt{b}}\cdot \frac{u'v-uv'}{v^2}$

$u=\sqrt{ax+b}-\sqrt{b}\\ u'=\frac{a}{2}(ax+b)^{-\frac{1}{2}}\\ v=\sqrt{ax+b}+\sqrt{b}\\ v'=\frac{a}{2}(ax+b)^{-\frac{1}{2}}$

                            v                         u'                  v                       u                   v'                          

$\large f'(x)=\frac{\sqrt{ax+b}+\sqrt{b}}{\sqrt{ax+b}-\sqrt{b}}\cdot \frac{\frac{a}{2}(ax+b)^{-\frac{1}{2}}(\sqrt{ax+b}+\sqrt{b})-(\sqrt{ax+b}-\sqrt{b})(\frac{a}{2}(ax+b)^{-\frac{1}{2}})}{(\sqrt{ax+b}+\sqrt{b})^2}$

                            u                                                            v²

$\large f'(x)=\frac{\sqrt{ax+b}+\sqrt{b}}{\sqrt{ax+b}-\sqrt{b}}\cdot \frac{a(ax+b)^{-\frac{1}{2}}(\sqrt{ax+b}+\sqrt{b}-\sqrt{ax+b} +\sqrt{b})}{2\cdot (\sqrt{ax+b}+\sqrt{b})^2}$

$\large f'(x)=\frac{1}{\sqrt{ax+b}-\sqrt{b}}\cdot \frac{a(\sqrt{ax+b}+\sqrt{b}-\sqrt{ax+b} +\sqrt{b})}{2\cdot (\sqrt{ax+b}+\sqrt{b})(\sqrt{ax+b})}$ 

$\large f'(x)=\frac{1}{x}\cdot \frac{\sqrt{b}}{\sqrt{ax+b}}$

$\large f'(x)=\frac{1}{x}\cdot\sqrt {\frac{b}{ax+b}}$

Gruß

  !