크레이지J의 탐구생활

21. $\frac{d}{dx}(ysiny = xsinx)$

$\begin{aligned} &\frac{d}{dx}(ysiny = xsinx)\\ &=y'siny+ycosyy'=sinx+xcosx\\ &=y'(siny+ycosy)=sinx+xcosx\\ &y'=\frac{sinx+xcosx}{siny+ycosy} \end{aligned}$

22. $\frac{d}{dx}(ln(x/y) = e^{xy^3})$

$\begin{aligned} &\frac{d}{dx}(ln(x/y) = e^{xy^3})\\ &=\frac{y}{x}(x/y)'=e^{xy^3}(xy^3)'\\ &=\frac{y}{x}\frac{y-xy'}{y^2}=e^{xy^3}(y^3+x3y^2y')\\ &\frac{y^2-xyy'}{xy^2}=y^3e^{xy^3}+3xy^2e^{xy^3}y'\\ &\frac{1}{x}-\frac{y'}{y}=y^3e^{xy^3}+3xy^2e^{xy^3}y'\\ &\frac{1}{x}-y^3e^{xy^3}=3xy^2e^{xy^3}y'+\frac{y'}{y}\\ &\frac{1-xy^3e^{xy^3}}{x}=(\frac{1+3xy^3e^{xy^3}}{y})y'\\ &y'=\frac{1-xy^3e^{xy^3}}{x}(\frac{y}{1+3xy^3e^{xy^3}})\\ &y'=\frac{y-xy^4e^{xy^3}}{x+3x^2y^3e^{xy^3}}\\ \end{aligned}$
Alt.
$\begin{aligned} &\frac{d}{dx}(ln(x/y) = e^{xy^3})\\ &\frac{d}{dx}(ln(x)-ln(y) = e^{xy^3})\\ &\frac{1}{x}-\frac{y'}{y}=e^{xy^3}(y^3+x3y^2 y')\\ &\text{same as upper} \end{aligned}$

23. $\frac{d}{dx}(x=sec(y))$

$\begin{aligned} &\frac{d}{dx}(x=sec(y))\\ &dx=sec(y)tan(y)dy\\ &\frac{dy}{dx}=\frac{1}{sec(y)tan(y)}\\ &(R.T. angle=y, a=1,h=x, o=\sqrt{x^2-1})\\ &\frac{dy}{dx}=\frac{1}{x\sqrt{x^2-1}}=(arctan(x))'\\ \end{aligned}$

24. $\frac{d}{dx}((x-y)^2 = sinx + siny )$

$\begin{aligned} &\frac{d}{dx}((x-y)^2 = sinx + siny )\\ &2(x-y)(1-y')=cosx+cosyy'\\ &=2(x-y)-2(x-y)y'=cosx+cosyy'\\ &=2(x-y)-cosx=cosyy'+2(x-y)y'\\ &y'(cosy+2(x-y))=2(x-y)-cosx\\ &y'=\frac{2(x-y)-cosx}{2(x-y)+cosy}\\ \end{aligned}$

25. $\frac{d}{dx}( x^y = y^x)$

$\begin{aligned} &\frac{d}{dx}( x^y = y^x)\\ &y ln x = x ln y \\ &y' lnx + y(1/x) =lny+x(1/y)y' \\ &y'(lnx-x(1/y))=ln y-y(1/x) \\ &y' = \frac{ln y-y(1/x)}{lnx-x(1/y)} \\ &= \frac{xy ln y-y^2}{ xylnx-x^2} \\ \end{aligned}$

26. $\frac{d}{dx}(arctan(x^2y) = x+y^3)$

$\begin{aligned} &\frac{d}{dx}(arctan(x^2y) = x+y^3)\\ &y=arctan x, x=tan y, dx=sec^2ydy\\ &R.T angle=y, a=1, o=x, h=sqrt(x^2+1)\\ &dy/dx = \frac{1}{sec^2y}=cos^2y=\frac{1}{1+x^2} \\ \\ &\frac{d}{dx}(arctan(x^2y) = x+y^3)\\ &\frac{1}{1+x^4y^2}(2xy+x^2y')=1+3y^2y'\\ &\frac{2xy}{1+x^4y^2}+\frac{x^2y'}{1+x^4y^2} =1+3y^2y'\\ &\frac{2xy}{1+x^4y^2}-1= 3y^2y'-\frac{x^2y'}{1+x^4y^2}\\ &\frac{2xy-1-x^4y^2}{1+x^4y^2}= (3y^2-\frac{x^2}{1+x^4y^2})y'\\ &\frac{2xy-1-x^4y^2}{1+x^4y^2}= \frac{3y^2+3x^4y^4-x^2}{1+x^4y^2}y'\\ &y'=\frac{2xy-1-x^4y^2}{3y^2+3x^4y^4-x^2}\\ &=\frac{x^4y^2-2xy+1}{-3x^4y^4+x^2-3y^2}\\ \end{aligned}$

27. $\frac{d}{dx}(x^2/(x^2-y^2) = 3y)$

$\begin{aligned} &\frac{d}{dx}(x^2/(x^2-y^2) = 3y)\\ &\frac{d}{dx}(x^2 = 3y(x^2-y^2)=3x^2y-3y^3)\\ &2x=6xy+3x^2y'-9y^2y'\\ &2x-6xy=y'(3x^2-9y^2)\\ &y'=\frac{2x-6xy}{3x^2-9y^2} \end{aligned}$

28. $\frac{d}{dx}(e^{x/y} = x + y^2)$

$\begin{aligned} &\frac{d}{dx}(e^{x/y} = x + y^2)\\ &\frac{d}{dx}(\frac{x}{y} = ln(x + y^2))\\ &\frac{y-xy'}{y^2} =\frac{1}{x + y^2}(1+2yy')\\ &\frac{1}{y} - \frac{1}{x + y^2}=\frac{xy'}{y^2}+\frac{2yy'}{x + y^2}\\ &y(x+y^2) - y^2=x(x+y^2)y'+2y^3y'\\ &y'=\frac{y^3-y^2+xy}{x^2+xy^2+2y^3}=\frac{y(x+y^2)-y^2}{x(x+y^2)+2y^3}\\ &=\frac{ye^{x/y}-y^2}{xe^{x/y}+2y^3} \end{aligned}$

29. $\frac{d}{dx}((x^2 + y^2 – 1)^3 = y)$

$\begin{aligned} &\frac{d}{dx}((x^2 + y^2 – 1)^3 = y)\\ &3(x^2+y^2-1)^2(2x+2yy')=y'\\ &3(x^2+y^2-1)^2(2x)+3(x^2+y^2-1)^2(2yy')=y'\\ &6x(x^2+y^2-1)^2+6y(x^2+y^2-1)^2y'=y'\\ &y'=\frac{6x(x^2+y^2-1)^2}{1-6y(x^2+y^2-1)^2}\\ \end{aligned}$

30. $\frac{d^2y}{dx^2} (9x^2 + y^2 = 9)$

$\begin{aligned} &\frac{d^2y}{dx^2} (9x^2 + y^2 = 9)\\ &\frac{d}{dx}(\frac{d}{dx} (9x^2 + y^2 = 9) )\\ &\frac{d}{dx} (9x^2 + y^2 = 9) \\ &18x+2yy'=0\\ &y'=-\frac{9x}{y} \\ &y'' = -9\frac{y-xy'}{y^2}=-9\frac{y-x(-9\frac{x}{y})}{y^2}\\ &=-9\frac{y+9x^2/y}{y^2}=-\frac{9}{y}-81\frac{x^2}{y^3}\\ &=-9\frac{y^2+9x^2}{y^3}=-\frac{81}{y^3} \end{aligned}$

Author: crazyj7@gmail.com