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How to Mathe Abi

04) Ableitungen (Analysis)

RK
von Rick K.

Mit welcher Formel berechnest Du die mittlere / durchschnittliche Änderungsrate einer Funktion f im Intervall


I=[a;b]I=[a;b]



mittlere / durchschnittliche Änderungsrate:


m=f(b)−f(a)b−a‾‾\underline{\underline{m=\frac{f(b)-f(a)}{b-a}}}


Berechne die mittlere Änderungsrate der Funktion:


f(x)=x2f(x)=x^2


im Intervall:


I=[0;4]I=[0;4]


Lösung:


m=42−024−0=164m=\frac{4^2-0^2}{4-0}=\frac{16}{4}


m=4‾‾\underline{\underline{m=4}}



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f(x)=x3f(x)=x^3


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I=[0;3]I=[0;3]


Tivmvwo


m=33−033−0=273m=\frac{3^3-0^3}{3-0}=\frac{27}{3}


m=9‾‾\underline{\underline{m=9}}



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f(x)=exf(x)=e^x


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I=[0;1]I=[0;1]



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m=e1−e01−0=e−11m=\frac{e^1-e^0}{1-0}=\frac{e-1}{1}


m=e−1‾‾\underline{\underline{m=e-1}}



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m=f′(a)‾‾\underline{\underline{m=f'(a)}}


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m=f′(a)‾‾\underline{\underline{m=f'(a)}}


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f(x)=x2+3xf(x)=x^2+3x


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f′(x)=2x+3‾‾\underline{\underline{f'(x)=2x+3}}


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f(x)=x2+3xf(x)=x^2+3x


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f′(x)=2x+3f'(x)=2x+3


m=f′(0)=2⋅0+3m=f'(0)=2\cdot 0 +3


m=3‾‾\underline{\underline{m=3}}


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f(x)=4⋅x2f(x)=4\cdot x^2


f′(x)=4⋅2⋅xf'(x)=4\cdot 2\cdot x


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f(x)=e⋅xf(x)=e\cdot x


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f′(x)=e‾‾\underline{\underline{f'(x)=e}}


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f(x)=ef(x)=e


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f′(x)=0‾‾\underline{\underline{f'(x)=0}}


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f(x)=x3+3x2+1f(x)=x^3+3x^2+1


Whpoyri


f′(x)=3x2+6x‾‾\underline{\underline{f'(x)=3x^2+6x}}


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f(x)=x4+ef(x)=x^4+e


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f′(x)=4x3‾‾\underline{\underline{f'(x)=4x^3}}


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f(x)=xnf(x)=x^n


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f′(x)=n⋅xn−1‾‾\underline{\underline{f'(x)=n\cdot x^{n-1}}}


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f(x)=g(x)+h(x)f(x)=g(x)+h(x)


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f′(x)=g′(x)+h′(x)‾‾\underline{\underline{f'(x)=g'(x)+h'(x)}}


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f(x)=ln⁡(x2+x)f(x)=\ln(x^2+x)


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f(x)=x2⋅exf(x)=x^2\cdot e^x


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f(x)=u(x)⋅v(x)f(x)=u(x)\cdot v(x)

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f(x)=ex2−1f(x)=e^{x^2-1}


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f(x)=g(h(x))f(x)=g(h(x))

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g(t)=(4t+1)⋅etg(t)=(4t+1)\cdot e^t


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m=f′(x0)m=f'(x_{0})


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g(x)=x4+3x2+1g(x)=x^4+3x^2+1


Xxddj hom


g(x)=e3x+1g(x)=e^{3x+1}


Pshybsn


g′(x)=3⋅e3x+1‾‾\underline{\underline{g'(x)=3\cdot e^{3x+1}}}


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f(x)=x3f(x)=x^3


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x0=0x_0=0


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f′(x0)=3⋅02f'(x_0)=3\cdot 0^2


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m=0‾‾\underline{\underline{m=0}}


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f(x)=x3f(x)=x^3


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x0=0x_0=0

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n(t)=(4t+1)2n(t)=(4t+1)^2


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n′(t)=4⋅2⋅(4t+1)n'(t)=4\cdot 2\cdot (4t+1)


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n′(t)=8⋅(4t+1)‾‾\underline{\underline{n'(t)=8\cdot (4t+1)}}


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f(x)=u(x)⋅v(x)f(x)=u(x)\cdot v(x)


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f′(x)=u′(x)⋅v(x)+v′(x)⋅u(x)‾‾\underline{\underline{f'(x)=u'(x)\cdot v(x) + v'(x)\cdot u(x)}}


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u′v+v′u‾‾\underline{\underline{u'v+v'u}}


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f(x)=ln⁡(x2+1)f(x)=\ln(x^2+1)


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h(x)=x⋅exh(x)=x\cdot e^x


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h′(x)=1⋅ex+ex⋅xh'(x)=1\cdot e^x + e^x\cdot x

msh vejboc


h′(x)=ex⋅(1+x)‾‾\underline{\underline{h'(x)=e^x\cdot (1+x)}}


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f(x)=ex2+2x−1f(x)=e^{x^2+2x-1}


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h(x)=x2+2x−1‾‾\underline{\underline{h(x)=x^2+2x-1}}


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f(x)=2x3+2x+1f(x)=2x^3+2x+1


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f′(x)=6x2+2f'(x)=6x^2+2


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f′′(x)=12x‾‾\underline{\underline{f''(x)=12x}}


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f(x)=x2f(x)=x^2


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x0=4x_0=4


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f(x)=ln⁡(x)f(x)=\ln(x)


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f′(x)=1x‾‾\underline{\underline{f'(x)=\frac{1}{x}}}


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f(x)=uvf(x)=\frac{u}{v}


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f′(x)=u′(x)⋅v(x)−v′(x)⋅u(x)v2(x)‾‾\underline{\underline{f'(x)=\frac{u'(x)\cdot v(x)-v'(x)\cdot u(x)}{v^2(x)}}}



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f(x)=x⋅ln⁡(x2−1)f(x)=x\cdot \ln(x^2-1)


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u(x)=xu(x)=x


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v(x)=ln(x2−1)v(x)=ln(x^2-1)


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f(x)=ln⁡(4x)f(x)=\ln(4x)


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f′(x)=44xf'(x)=\frac{4}{4x}


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f′(x)=1x‾‾\underline{\underline{f'(x)=\frac{1}{x}}}


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f(x)=2⋅ln⁡(3x)f(x)=2\cdot \ln(3x)


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f′(x)=2⋅13x⋅3f'(x)=2\cdot \frac{1}{3x}\cdot 3


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f′(x)=2x‾‾\underline{\underline{f'(x)=\frac{2}{x}}}


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f(x)=xf(x)=\sqrt{x}


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f(x)=x12‾‾\underline{\underline{f(x)=x^{\frac{1}{2}}}}


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n=12n=\frac{1}{2}


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f(x)=xf(x)=\sqrt{x}


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f(x)=12x‾‾\underline{\underline{f(x)=\frac{1}{2\sqrt{x}}}}


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f(x)=ex2+xf(x)=e^{x^2+x}


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f′(x)=(2x+1)⋅ex2+x‾‾\underline{\underline{f'(x)=(2x+1)\cdot e^{x^2+x}}}


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f(x)=1x5f(x)=\frac{1}{x^5}


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f(x)=x−5‾‾\underline{\underline{f(x)=x^{-5}}}


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n=−5n=-5


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g(x)=2x14g(x)=\frac{2}{x^{14}}


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g(x)=2⋅x−14g(x)=2\cdot x^{-14}


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n=−14n=-14


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f(x)=ln⁡(g(x))f(x)=\ln(g(x))


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f′(x)=g′(x)g(x)‾‾\underline{\underline{f'(x)=\frac{g'(x)}{g(x)}}}


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f(x)=ln(4x2+x)f(x)=ln(4x^2+x)


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f′(x)=8x+14x2+x‾‾\underline{\underline{f'(x)=\frac{8x+1}{4x^2+x}}}


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f(x)=ex2−1f(x)=e^{x^2-1}


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g(x)=e...g(x)=e^{...}


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f(x)=x⋅ln(x2+1)f(x)=x\cdot ln(x^2+1)


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f(t)=ln⁡(2x)+1f(t)=\ln(2x)+1


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f(t)=0‾‾\underline{\underline{f(t)=0}}


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f(x)=1x4f(x)=\frac{1}{x^4}


Tfs nsv Gjkillgeklzeldryzt ceohe


f′(x)=−4x5‾‾\underline{\underline{f'(x)=\frac{-4}{x^5}}}


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f(t)=1t3f(t)=\frac{1}{t^3}


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f′(t)=−3t4‾‾\underline{\underline{f'(t)=\frac{-3}{t^4}}}


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f(s)=4⋅e2s+1f(s)=4\cdot e^{2s+1}


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f′(s)=8⋅e2s+1‾‾\underline{\underline{f'(s)=8\cdot e^{2s+1}}}


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f(x)=(x+1)⋅(x2−4)f(x)=(x+1)\cdot (x^2-4)


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f(x)=eh(x)f(x)=e^{h(x)}


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f′(x)=h′(x)⋅eh(x)‾‾\underline{\underline{f'(x)=h'(x)\cdot e^{h(x)}}}


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fa(x)=eax+bf_a(x)=e^{ax+b}


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fa′(x)=a⋅eax+b‾‾\underline{\underline{f_a'(x)=a\cdot e^{ax+b}}}


Ncble psv


f(x)=x+1f(x)=\sqrt{x+1}


Vxmczbm


f′(x)=12⋅(x+1)−12‾‾\underline{\underline{f'(x)=\frac{1}{2}\cdot (x+1)^{-\frac {1}{2}}}}


Mjjpz ocq


q(x)=13⋅ln⁡(x)q(x)=\frac{1}{3}\cdot \ln(x)


Uvkauoi


q′(x)=13x‾‾\underline{\underline{q'(x)=\frac{1}{3x}}}


Jvxwn mvd


f(x)=exxf(x)=\frac{e^x}{x}


Quv Scscnhyke zlw zwqxzu Uizpfwsra qyoslbo


f′(x)=ex⋅x−1⋅exx2f'(x)=\frac{e^x\cdot x - 1\cdot e^x}{x^2}


jzjhn


f′(x)=ex⋅(x−1)x2‾‾\underline{\underline{f'(x)=\frac{e^x\cdot (x-1)}{x^2}}}


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