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Question Number 192338 by josemate19 last updated on 15/May/23

$$y=\left(\frac{\left(\underset{h\to 0}{\lim }\frac{{(x+h)}^3-x^3}{h}\right)\left(\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{x^{n+1}}{n+1}\right)}{{\int }_0^{x}lntdt}\right)$$$$$$$$\frac{dy}{dx}?$$

Answered by aleks041103 last updated on 15/May/23

$$\underset{h\to 0}{lim}\frac{{(x+h)}^3-x^3}{h}={\left(x^3\right)}^{\prime }=3x^2$$$$\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{x^{n+1}}{n+1}=\underset{n=0}{\sum ^{\mathrm{\infty }}}{\int }_0^x{t}^{n}dt={\int }_0^x\left(\underset{n=0}{\sum ^{\mathrm{\infty }}}{t}^n\right)dt=$$$$={\int }_0^x\frac{1}{1-t}dt={\int }_{1-x}^1\frac{dt}{t}=-ln(1-x)$$$${\int }_0^{x}ln\left(t\right)dt={\left(tln\left(t\right)\right)}_0^x-{\int }_0^{x}td\left(ln\left(t\right)\right)=$$$$=xln\left(x\right)-x$$$$\Rightarrow y=\frac{3x^{2}ln(1-x)}{x(1-ln\left(x\right))}=\frac{3xln(1-x)}{1-ln\left(x\right)}$$$$y^{\prime }=\frac{(-\frac{3x}{1-x}+3ln(1-x))(1-ln\left(x\right))+3ln(1-x)}{{(1-ln\left(x\right))}^2}$$

Answered by mehdee42 last updated on 15/May/23

$$p1){lim}_{h\to 0}\frac{{(x+h)}^3-x^3}{h}=3x^2$$$$p2)\underset{0}{\sum ^{\mathrm{\infty }}}\frac{x^{n+1}}{n+1}=-ln(1-x)taylorexpantion$$$$p3){\int }_0^{x}lntdt={lim}_{h\to 0}[tlnt-t]\underset{h}{\stackrel{x}{}}=xlnx-x$$$$\Rightarrow y=\frac{-3xln(1-x)}{lnx-1}$$$$\Rightarrow y^{\prime }=\frac{(-3ln(1-x)+3x\times \frac{1}{1-x})(lnx-1)+3ln(1-x)}{{(lnx-1)}^2}$$

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