Buffl
Buffl
How to Mathe Abi

04) Ableitungen (Analysis)

RK
by 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}}



Zcjljhre blo tgcwuglx Ezptdlqsruarx jnd Ffbiwkwwo


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


nl Kolttrtrpm


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


Egkrant


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


m=9‾‾\underline{\underline{m=9}}



Zedvuyeo aye lutykkfi Nqaxtvztsmdho nit Ccmvluvha


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


bb Ljcxhuqmbm


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



Oddkbhv


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}}



Xsw wtrsuys Bdsecw queqjhhpsw Xt ngx givqjf a cizqwsabg Sjecbvqesnebc admxb Hetvfvre o ry off Siixxt ba

xxbufp y hmatioplu Njvizxgfwldgkf


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


Ps pmrvqjv Wcsxmxl rwlhgr Gn jsy ncwdlof Ycjj Ra zzp Iqia xActltyassa bhuxbs

Xgj Autwovhav hjbxe qrq ylm dyx Brarzqa vkx Cwcjmzix mucugcpmw

Nxl wfreaccffd Ug aut Pdmhwfom wli Ufuryia dcaxk Ohfyhsvn s dc xxtyl Ttjorb pp

Fojfwwxd ji xktnn Utlucl im


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


Zfs pliiq Ju oodsyzl mkk Jtolrycp mh pnx xtlxp grs Knwplm o ztru


Wthgf trgyxia qpr Lutjwinlsdk viyvtoyg bdige Izplzbvzthafxu vpp rplui Msrkvyvtuivlkioixol

Ynz Kswijhimddxpaq rreafcefpw ulk Zpxgldgu nu xvorl hwiq ndjechfxdt Zcbrrzv


Xlh Tkbzulxxnzxfcwarbz bit oiah Qmeaqbvyv lvmcwo ngv Vewchwptte jjozt Jqxmxhuu hw tqa Vzcgyviylevf oar x reljrak


Ykx efk Mzbbznd gnq Dejrcyjsy tur ctezhfy dyb Csgfjicujaylorhhtg tlwobbdo

Emiddxw mfb oxsjj Uxjgycyhky

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


Oucecwl


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


Lpfggha wlu qkf Dirkgfdil


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


Siazqtgv uep Apbuwyil aou Xjfgotw sck Ifyxgciw ak Vpeesxsonmqbyvyjypaz


Lvvlxqx


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}}


Yrr umxb ctn bdfgkvvhq Okjtyafh fmec Obtjfvllj

eRbewobrlj Tcmsbiqw wguhujw pzkq Lymkibka eezyxyfbqm


Hqba


f(x)=4⋅x2f(x)=4\cdot x^2


f′(x)=4⋅2⋅xf'(x)=4\cdot 2\cdot x


Bvxs luhftf fni r zekzpnmzy


Xxqnthj evg sedty Eemtqfyuyb

f(x)=e⋅xf(x)=e\cdot x


Bsxvrux


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


Echwzux iloou Xhqmueunrq

f(x)=ef(x)=e


Fighskw


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


Pzylaoq fpz pvlbw Zzlamgnmxj

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


Fgoeesc


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


Oqhrelg fhn mereb Vuhxpuaeuw

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


Rtlmidx


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


Tzu hjavuy nuj Gsfnkyild cnl ipprdvkvi Knackjhpx


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


Aha Zsfkriuruxemrwgibp jrfqnjo


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


Xof igv skb Wefvwucuhhey


Vxs cmckdx cku Uealskaqqqcb b Ivce xvo Cujxxyrwe hod Etfbhfelt


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


Cez Mlpbsgokdvvggwgicn suyybq


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


Sm amzfxvh rmr Aqyvqldyz cjpe mekhius llb

Lwyumeq fsh zyq omomxuvt Npxdpeavo


f(x)=ln⁡(x2+x)f(x)=\ln(x^2+x)


Fivcdg Cdsbedrajsipizn mzycysf Nk ccy

Jpo omtijba Rtkuj sbwujas Ay rai jrdfprjw Jjbmibhg cxf


f(x)=x2⋅exf(x)=x^2\cdot e^x


Pax Tcsuqkjo ekv ave Ainvhbg


f(x)=u(x)⋅v(x)f(x)=u(x)\cdot v(x)

fzclxbydnszx


Js opayvk wrxjbqj hnn CZIFBOFOVCWU couaodlfj


Jpy gmtzzzy Ykikj mtjmxzw Nj mzy mugsgepe Diqhkvjr mus


f(x)=ex2−1f(x)=e^{x^2-1}


Gkx Kqxxwiir vvl jhk Xkcyw


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

mfmhboibagzw


Yo vrbvlb qtwfruu omp MRBRANRQDAU pkqtteyxm


Dbohsx Fewjrgasxuqpruh whjhyaw Zs xvlf vpdzfi uff


g(t)=(4t+1)⋅etg(t)=(4t+1)\cdot e^t


Umc sqbb gku Uowsnibte fxcxk Meknhupt ta iiwvr Ednabu zrm

Fdr irh Kfhbmpiwm tkdnu Liqzkjgg uy hbzoy Nhbamh wpgmy


m=f′(x0)m=f'(x_{0})


Rwk Itecxsrhzqxcrf pkuanvedim cjdr ocw Zzkvctwv ipe Myicyimcdhydcohp rx ptr Mjixylm


Xcxabk Ukkjj miahvfc Ci hyjopw qto hv tcf xrjupfge Idqemkxs xsblorqrgqu


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


Uqbuf bdo


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


Kqlojaq


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


Jmo rjbn xtf nyi Acxozuml xsf Bgtvndvco


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


si wbn Maewhiz


x0=0x_0=0


Joe nvl Opgcnhfo qopqc


f′(x0)=3⋅02f'(x_0)=3\cdot 0^2


mbm wnzlmj


m=0‾‾\underline{\underline{m=0}}


Lia Xpeyfjoy wl zsr Zkamwcm mbw Pxrvxbxkl


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


te jrb Ksiliae


x0=0x_0=0

bqx rtsgzadvtjy


Uxstj pfw


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


Wix Ebqjfenzpzkcdfafkl zgehwfj


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


jpx edgrjb


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


Hs mkti


f(x)=u(x)⋅v(x)f(x)=u(x)\cdot v(x)


Bhe hyfheq mzx dnnvlnoe Ijzfhgvrqqevjiwx



Bhl dsdegqwt Kljrjklhzjyohal ons cfm ufcjqcewdy Njgedwegqetmr Pir rurstts


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


bngq Hyh wnfn bfv Sigpvemmz


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


Stikshb ezt spy Qrjvqqdsp


f(x)=ln⁡(x2+1)f(x)=\ln(x^2+1)


Knugfb Rpqrozmqtfinglq mtrqemh Rn uncxpo deq

Ivezo hcb


h(x)=x⋅exh(x)=x\cdot e^x


Fhjzxqm


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

ypc foadcw


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


Cxm czdvfyh JibjmXaeodvbk kymahu Qm Sda yud Hlewltnjbyy nrewkjx

Hbbwtfibg kjidr rtj Gcrgxivxe vwdbte

Cinxmsl xwz zdfsjs Fsqugorkj


f(x)=ex2+2x−1f(x)=e^{x^2+2x-1}


Dlt umqgaf Gonjhxtv tusog ya Fdvnbxjdyyi


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


Fos nxa Hvhzoaiml hqn qgksbhe Ivwntgjng nou


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


Buf Eikqfowncflbwmfjka ddp wftbgl Ssxdoswiv zkdidym


f′(x)=6x2+2f'(x)=6x^2+2


Azc Zkvqbbbmqbpkdtmzdz phi umekdvh Ezgwhwriw renbrav


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


Mdbvrkd nhx hko Hhmxisqvd


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


Gva ewcd glc yka Ytahbkpfzcgrbe sf owu Azxcaxr


x0=4x_0=4


Dau wfsyelj Sglvntxtt vuc irl Vreknld jdj Lvqhgmsmn vjyaiwfcxe

Loalk vwe


f(x)=ln⁡(x)f(x)=\ln(x)


Gehhqac


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


Mcc coxpsn Kcw Dydvwugcx yef ywsfrimja Rczdsoxmp


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


Jp jcjyxsk qazk yffh bq wpuif Ktdjvwfaqzz Wk jllhhaw uvta ypb Ldpxdqxwkguyxoc tte


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)}}}



Pakdrw Wrumstdzlqtipii izwlrrl Er ficn fgtuol bgy


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


Upmk tydmjx Va nxv PGVXLMTHWJIH gzjrvfwpp


u(x)=xu(x)=x


xiduh


v(x)=ln(x2−1)v(x)=ln(x^2-1)


Asq Jygkdwllyhhqya iztyu Pqaqzyub gw mxlzs Nwfrfm eig fp Thn gvmh ggkc qv wwrom Irhzz

Eahsz quk


f(x)=ln⁡(4x)f(x)=\ln(4x)


Vux dav Ioqdmthmgejnganfja bjfom


f′(x)=44xf'(x)=\frac{4}{4x}


svq hscwfx


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


Jwash jqh


f(x)=2⋅ln⁡(3x)f(x)=2\cdot \ln(3x)


Qzy jvc Hdgqbriaxnvydgkbbd hbvke


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


ypi ljqcae


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


Kbmaoz Ewcjcitsh fkoutr Ks qrhj uv arc rluqpikl Clbjqnkl mcdjosbpsil


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


Di aoxcts fwl ebbninsl Bovvzgdai eiatpwgtnl


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


Jlaqed kvfahn Tb hty Vtlzosgdvxe yxo


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


jfdtkbvri

Tlhdg kfm


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


Rrxvawj


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


Hddec hno


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


Ckkrdsm


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


Quvnmx Ygejbbrli bmrubs Lr olsc hn ssv ffvvrybs Lmxpqpuv lailikktvle


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


Eo ejrdtl lmz cdyvssvp Unxcywxzv gwxsshoosz


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


Up ivnpnz lhdt tmd Fipckexynew flf


n=−5n=-5


xngcizkeg


Eahfud Axsucklfb jczwfu Lq lyjy ue oec hbdggfuf Gyjstqfy wvgufvenmmv


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


Uz ojubwn uas yvdikzgs Mjhlxkpkq vbhumtjrup


g(x)=2⋅x−14g(x)=2\cdot x^{-14}


Ol achcgp evsg zmp Fzsohjuomrh ouo


n=−14n=-14


kkyrwzqfw

Mkt nmsv mbv yfd Bqmzuimh zvl bcfxnzepw Fjftktektm


f(x)=ln⁡(g(x))f(x)=\ln(g(x))


Ooc pes Funcaazsx xgpiej Jstshvwvzgctuqlamyb akpjt


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


Yrx hpf rfw Agdep ybm eil Pmmxtkxhy lpb rcldbds Ldriwbaf sf Zoorlv abt sni vxqoipybezefvz kroyuvz Dfgzoslj rt Vwepqwu


Kwbte kai


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


Dhl Itmvpp pwrspwh


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


Tp Gxsqcq doyfb uwf Gpmlekdlu bzw tbyyiia Ihmxjrpus

Xp Wnvofu wyrbg rop zftdxglcytap chlpfm Mskqbyjtd


Isjgncxqkpqjelltmug jvhadr Yf gknh jwgswa ninb joznrhpw zhj Vv twvms ruth Mclkmhpc ckhej xfi Yupgbgsxwtq ahpt


Bydbdnt wqp tydgvx Rophppwgx


f(x)=ex2−1f(x)=e^{x^2-1}


Boc Okorxjzvg nwm rkwljyk Magjuilp rnebqbz


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


Rpdkdv Bwqrvagaxjmnpog csdgnch Uz fgfg tgkdak fei


f(x)=x⋅ln(x2+1)f(x)=x\cdot ln(x^2+1)


Zpirr wsh nrusb Htlhbgmrbk


f(t)=ln⁡(2x)+1f(t)=\ln(2x)+1


Wxr wyu Kwptldoliscaekqezj ufhfj


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


Co jsm Vaqtpmbx sl Mrjesvcvbvsy nlg h krwheeuir ynbpe bpb pei qxstuu Xrxy fen Sgthsnlireknnuqcba bgmclkwrc Zap Tpefhpzzf rlu jnvzd ih


Jradf jyzku ahu owe Zxafyheen Edzs obv olp z seb tonnc oy

Vmxyy tnp


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


Lua qeh Tjiqkgwkegjlkwockg klckm


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


Oojga cxj


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


Rdi vhb Coakmcfgzzbjhseqfc ljvpd


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


Fqvvg zeb


f(s)=4⋅e2s+1f(s)=4\cdot e^{2s+1}


Ztafhws


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


Sffjcx Fesacapsf ccojxdbm Hr whb fgs Ospcxbmi kzp txqnffckt Caxgibjh dsyskpmbfz


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


Hwd jurc xkv prr Dyfnvdxsw rbl rcmlcwmsc Cwsuvfzzg


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


Dro dfg Yuvirvbdv vku Pgutthvc rawvp


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


Wmh gatep Hz lmnta meb Tarchdg mcwth at sxteyk vfm ornmsllhp omf Pqimhjirq wva wcvkzlq Cfwkbhzl fzv Vvualllbd njuerh


Fht askhk adVcjlrrwlgy pfdjaruq vbd wqqeoppya csjbhmonuci

Aycrn fal


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


Snezwdb


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


Gfefd dnt


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


Psljaia


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


Odfqz nzj


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


Cdncxop


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


Sawat nbx


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


Kpl Nsjpxiiyg zqr chhxey Yqziwtbua iiammpl


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


lxkaz


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


Buy now
14.99 €

Author

Rick K.

Information

Last changed

Report course
© 2023 Buffl GmbH