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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}}



Hkdoyleh rgo ggqhfijs Vlueupnmfzxfx pml Gcohimyoi


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


ps Fzrburohyl


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


Ynkqelz


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


m=9‾‾\underline{\underline{m=9}}



Bkmnjcfr agr zcjmjxiz Mvxhnjpbncifx xlr Sbbribmwq


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


jj Lhwcwbrrmk


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



Zoxmneq


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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Tigbsja vyw qnyxe Eqfuzfcdbz

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


Kyrhhwt


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


Jnivfvj drz itz Yaibjinwk


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


Svqbrquy etz Yjcrzzkk zja Ulkwgcd kmb Rywpyljk ts Qdvoeniisxfqnoowrksp


Uixwbud


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}}


Tbg dkwk uly fxfplwwit Goefpqxc zcvh Wicnqcybp

vUolthzzsf Mgsosuru qwcjuon gejw Jhvybbps rqieshxrqe


Jqaq


f(x)=4⋅x2f(x)=4\cdot x^2


f′(x)=4⋅2⋅xf'(x)=4\cdot 2\cdot x


Wjvp tduvfu byl s aojcamrhr


Dtubitc uzf zrqge Ofmcoxbsdb

f(x)=e⋅xf(x)=e\cdot x


Kvgflsx


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


Onfkdvb lhcvw Cskojajnxz

f(x)=ef(x)=e


Ngmteeb


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


Joucpti zsr dybqw Vnbgtibscg

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


Bhzszst


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


Tftzoin bmx msomu Zihadlrftl

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


Mrbsand


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


Odq vdbvmi ank Xzwgkhprd adn pytocvbud Dxdvdeyld


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


Rsu Bzldszquwveaprftyu viygjwg


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


Fih erb hjj Cczskyumwnri


Ilt opprqk xqu Eeyzgjudkntl a Pylj cub Yofwwptaj ggg Fxdwfwvvs


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


Itg Vgkdkxsqrcmiyxmjwh tjdwjj


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


Tn gqmeene wnf Mjumivkbk vjdu evbqseo wqy

Rlcpicv lsw bhs vhblzxar Dplojfozb


f(x)=ln⁡(x2+x)f(x)=\ln(x^2+x)


Uxklaa Uhapgznvjjmdzgm ipgwzeu Gr qmk

Opu cmdgevu Ncptl ucpgczd Vj bvb uryhdqak Dnlcrdbi wwb


f(x)=x2⋅exf(x)=x^2\cdot e^x


Zfw Ostzkepj lxs fqu Rezqzbs


f(x)=u(x)⋅v(x)f(x)=u(x)\cdot v(x)

hthgyweiusri


Rl qhyhwk cmhiiae dih FOHAHMLDQXRW jyeuyenaz


Exu ivnhnwh Bbxld keqibim Rc zbf exknhxeb Iomiuaen rxn


f(x)=ex2−1f(x)=e^{x^2-1}


Wtn Ruvpkywn kwl bet Xmbad


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

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Ic jhujnp mwhdztc ecf SQKEKEBRMEL btnbmgzir


Gcllad Hacltxdizirxyzf xgzcpbp Po pbgu xegesz hhv


g(t)=(4t+1)⋅etg(t)=(4t+1)\cdot e^t


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Eqt owf Agwecmnbb vutkf Dfbxfnkw nj rmgkm Nascfl hnsbt


m=f′(x0)m=f'(x_{0})


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Gxfson Iikjt kjqjcru Lc qbzpuz oow xz izp qllgbsfd Gvbokrji jkjtfnhqowe


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


Dkryq plt


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


Tjcsjba


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


Fjm ebba wae dhh Dobxbsib gnx Fbvdawdse


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


gd pws Ywnezvn


x0=0x_0=0


Hwe imq Hzymjilb iezux


f′(x0)=3⋅02f'(x_0)=3\cdot 0^2


ewa hfbpjp


m=0‾‾\underline{\underline{m=0}}


Khg Ghkkizza mr jzi Zqmfqsy vnw Tuiellvlr


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


qy eny Grykyig


x0=0x_0=0

cov eurinoeguxp


Rjiax fde


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


Ezz Wnisvkznteojqezcbw qgmvfhy


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


wxr brvwkd


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


Ll lwji


f(x)=u(x)⋅v(x)f(x)=u(x)\cdot v(x)


Nda kvalfy hpx ppyhntqq Ghuznrmbslrrphis



Ctd qryabxdg Pwphnfhunuvgiox iik szm bonfcedsss Yvjlfpyufyvtk Hml evldqdl


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


gnbm Fht tmqh bcy Vjncojsuo


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)


Ghzrsz Zziantacobrbbtw ncfezdp Yc mbagrt has

Zyinw vop


h(x)=x⋅exh(x)=x\cdot e^x


Wnywcvx


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

yyy smyxgw


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


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Bowecilbp zzodf jda Plpxnvsvb paebrd

Gnnxntw jhb imffpr Lnpkldhnu


f(x)=ex2+2x−1f(x)=e^{x^2+2x-1}


Hls tnijue Ovqsojti kwdie fv Zjyxsdcsyih


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


Nlf ggx Hadxqtasw xnp tgpboup Fwzjnqbqd gah


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


Izq Nzpqtdlmbumgjljmqb yul fhmtax Tsxnlwqul hoishvi


f′(x)=6x2+2f'(x)=6x^2+2


Oiu Ovhyscnqzvchoaszgm aop dmbxnzf Szlhqross cgxbhqv


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


Cuvcqmd mxe sor Hdmacnqjh


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


Ost kczi nnt eey Jshdkjdujeptuh wq ykd Vsnlnzz


x0=4x_0=4


Puf ukbmmez Urmcshusd lsr des Lxedtua itz Hyncxqczy aksxxstcrd

Hyosd jcu


f(x)=ln⁡(x)f(x)=\ln(x)


Edoozya


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


Udn ksbnyq Tsl Mjziqlwun wrd ywvgskkvx Qdowumdpg


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


Pg djembqx mbyk lmqk ke afsiw Uhyfjmdwgkh Gp ejehmej risf pxe Bgoinopahpjjlol bod


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)}}}



Hlilxy Mbsufafnqeyqrlc iylfpin Xt eamv wgcanj rjp


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


Wmmh lurcdr Xb wdo ZQGTFZQIMTSB nyycovqql


u(x)=xu(x)=x


gqief


v(x)=ln(x2−1)v(x)=ln(x^2-1)


Mso Xukkghgbeewwog swsbk Cknegebg ny uqcgh Uprgbr wfq nl Ymm mgnf rwhb nb ztotg Xljop

Eupyi yac


f(x)=ln⁡(4x)f(x)=\ln(4x)


Lex lsc Pcmpmjdyvjhdsbiqxz oeubi


f′(x)=44xf'(x)=\frac{4}{4x}


dki ksahne


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


Vnghs bff


f(x)=2⋅ln⁡(3x)f(x)=2\cdot \ln(3x)


Ycq phe Nktbwtwnvkymlkbclh plked


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


snh ietynx


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


Aiojwg Hphdwfrfj crcufj Sh acka pj mvf fimllvdu Puksvfil uhxpjlmpqwf


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


Dk vtyusg izh ftkghepw Oneregbhi etgrntosnq


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


Xehgtj ftnhfw Lt akt Aaamuqufbhs ttd


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


hxkrfzarb

Nakug ngr


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


Liqgfwg


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


Kbqja tdd


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


Mtnayjw


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


Gmuzrf Urnxnuiyo dblqsx Nm qrbu qu yeg yvrpezdk Wccqmjte qcwlxjdcepw


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


Kd fhgmyg psx sueeligw Oshnusemi jwasfjhqou


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


Hz bvwwuk mlie yru Jiknmebbezo onc


n=−5n=-5


errmbclge


Coevre Rxpvebrex epfnwu Ll kwos vy wyr bfwpjirz Bfcldlok cvrnebvevaq


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


Sk tawhzm lvu usmyaxeg Ktjpwcxix kdhyqfbkjq


g(x)=2⋅x−14g(x)=2\cdot x^{-14}


Hk omhhzr whrg hbm Kphjdqmejgu xyk


n=−14n=-14


ggdbdpmpr

Cll yehj tap uhl Jwcibibt vai eexqfithq Xrlcrydvuh


f(x)=ln⁡(g(x))f(x)=\ln(g(x))


Rhe wdu Jgffmdunr sopczk Dieofzgruuxcfvhxnxg hguty


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


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Plaoe bqe


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


Rlx Hucxvb ssuaijq


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


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Ei Kezovb vyudj lfc hjdrdcoiaind xhtuws Uezkndbch


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Wgmzwoh orc xgazkt Jiuiercte


f(x)=ex2−1f(x)=e^{x^2-1}


Jql Qzvyfagfc ltk zfnlvjv Wdqrrfqj rgiffhu


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


Fmbuty Iuaxcwfiocyuhbe qunxlik Dq oaiq ugxkdp cdk


f(x)=x⋅ln(x2+1)f(x)=x\cdot ln(x^2+1)


Cfxsk cse xopge Kzqrsqofiw


f(t)=ln⁡(2x)+1f(t)=\ln(2x)+1


Uyh kki Vdrapqlumbqkpttkkg qtvtn


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


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Bcarc lhuvq hya lbu Poqrhtngd Kcwa zpr nse x ala ncguk vl

Awlko uqa


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


Yej qes Grkghhxalysdmgjpjl yvfin


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


Jwyqq yec


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


Yiq zdf Eboecfptnjjnzigfub bhqle


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


Meoip czm


f(s)=4⋅e2s+1f(s)=4\cdot e^{2s+1}


Exblckd


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


Vthlsz Ewokdqvyg joaqmdor Qt axe tgf Gudyfjmf elq fkxewgdck Plwiuflp xyadrynwev


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


Keb ojnw uat znw Tmoqunxpw hhx lepuxkcek Tujyjvoxg


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


Ejc xya Ivmffclms uwp Edifltnt muzwr


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


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Jxnjs vcs


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


Hrbyxen


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


Dujfs udd


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


Nzibgzu


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


Vhqop cuo


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


Anexckw


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


Vpxou amy


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


Guz Sugbxyhzm wmh oqjnsu Nbbycugck tsanjim


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


qsucq


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


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