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Question Number 192338 by josemate19 last updated on 15/May/23
y= ((((lim_(h→0) (((x+h)^3 −x^3 )/h))(Σ_(n=0) ^∞ (x^(n+1) /(n+1))))/(∫_0 ^( x) lnt dt))) (dy/dx)?
$${y}=\:\left(\frac{\left(\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left({x}+{h}\right)^{\mathrm{3}} −{x}^{\mathrm{3}} }{{h}}\right)\left(\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{x}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}\right)}{\int_{\mathrm{0}} ^{\:{x}} {lnt}\:{dt}}\right) \\ $$$$ \\ $$$$\frac{{dy}}{{dx}}? \\ $$
Answered by aleks041103 last updated on 15/May/23
lim_(h→0) (((x+h)^3 −x^3 )/h)=(x^3 )′=3x^2 Σ_(n=0) ^∞ (x^(n+1) /(n+1))=Σ_(n=0) ^∞ ∫_0 ^( x) t^n dt=∫_0 ^( x) (Σ_(n=0) ^∞ t^n )dt= =∫_0 ^( x) (1/(1−t))dt = ∫_(1−x) ^( 1) (dt/t)=−ln(1−x) ∫_0 ^( x) ln(t)dt = (t ln(t))_0 ^x −∫_0 ^( x) td(ln(t))= =xln(x)−x ⇒y=((3x^2 ln(1−x))/(x(1−ln(x))))=((3xln(1−x))/(1−ln(x))) y′=(((−((3x)/(1−x))+3ln(1−x))(1−ln(x))+3ln(1−x))/((1−ln(x))^2 ))
$$\underset{{h}\rightarrow\mathrm{0}} {{lim}}\frac{\left({x}+{h}\right)^{\mathrm{3}} −{x}^{\mathrm{3}} }{{h}}=\left({x}^{\mathrm{3}} \right)'=\mathrm{3}{x}^{\mathrm{2}} \\ $$$$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{x}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\:{x}} {t}^{{n}} {dt}=\int_{\mathrm{0}} ^{\:{x}} \left(\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{t}^{{n}} \right){dt}= \\ $$$$=\int_{\mathrm{0}} ^{\:{x}} \:\frac{\mathrm{1}}{\mathrm{1}−{t}}{dt}\:=\:\int_{\mathrm{1}−{x}} ^{\:\mathrm{1}} \frac{{dt}}{{t}}=−{ln}\left(\mathrm{1}−{x}\right) \\ $$$$\int_{\mathrm{0}} ^{\:{x}} {ln}\left({t}\right){dt}\:=\:\left({t}\:{ln}\left({t}\right)\right)_{\mathrm{0}} ^{{x}} −\int_{\mathrm{0}} ^{\:{x}} {td}\left({ln}\left({t}\right)\right)= \\ $$$$={xln}\left({x}\right)−{x} \\ $$$$\Rightarrow{y}=\frac{\mathrm{3}{x}^{\mathrm{2}} {ln}\left(\mathrm{1}−{x}\right)}{{x}\left(\mathrm{1}−{ln}\left({x}\right)\right)}=\frac{\mathrm{3}{xln}\left(\mathrm{1}−{x}\right)}{\mathrm{1}−{ln}\left({x}\right)} \\ $$$${y}'=\frac{\left(−\frac{\mathrm{3}{x}}{\mathrm{1}−{x}}+\mathrm{3}{ln}\left(\mathrm{1}−{x}\right)\right)\left(\mathrm{1}−{ln}\left({x}\right)\right)+\mathrm{3}{ln}\left(\mathrm{1}−{x}\right)}{\left(\mathrm{1}−{ln}\left({x}\right)\right)^{\mathrm{2}} } \\ $$
Answered by mehdee42 last updated on 15/May/23
p1)lim_(h→0) (((x+h)^3 −x^3 )/h)=3x^2 p2)Σ_0 ^∞ (x^(n+1) /(n+1))=−ln(1−x) taylor expantion p3)∫_0 ^x lntdt=lim_(h→0) [tlnt−t] _h ^x =xlnx−x ⇒y=((−3xln(1−x))/(lnx −1)) ⇒y′=(((−3ln(1−x)+3x×(1/(1−x)))(lnx −1)+3ln(1−x))/((lnx −1)^2 ))
$$\left.{p}\mathrm{1}\right){lim}_{{h}\rightarrow\mathrm{0}} \frac{\left({x}+{h}\right)^{\mathrm{3}} −{x}^{\mathrm{3}} }{{h}}=\mathrm{3}{x}^{\mathrm{2}} \\ $$$$\left.{p}\mathrm{2}\right)\underset{\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{{x}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}=−{ln}\left(\mathrm{1}−{x}\right)\:\:\:\:\:\:\:\:{taylor}\:\:{expantion} \\ $$$$\left.{p}\mathrm{3}\right)\int_{\mathrm{0}} ^{{x}} {lntdt}={lim}_{{h}\rightarrow\mathrm{0}} \left[{tlnt}−{t}\right]\underset{{h}} {\overset{{x}} {\:}}={xlnx}−{x} \\ $$$$\Rightarrow{y}=\frac{−\mathrm{3}{xln}\left(\mathrm{1}−{x}\right)}{{lnx}\:−\mathrm{1}} \\ $$$$\Rightarrow{y}'=\frac{\left(−\mathrm{3}{ln}\left(\mathrm{1}−{x}\right)+\mathrm{3}{x}×\frac{\mathrm{1}}{\mathrm{1}−{x}}\right)\left({lnx}\:−\mathrm{1}\right)+\mathrm{3}{ln}\left(\mathrm{1}−{x}\right)}{\left({lnx}\:−\mathrm{1}\right)^{\mathrm{2}} } \\ $$

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